Lemma. Every open cover has a $\sigma$-locally finite refinement which is also an open cover.

Proof. Using the axiom of choice, let $\mathcal{U} = (U_\alpha)_{\alpha < \xi}$ be an open cover indexed by some ordinal $\xi$. Let $n \in \N$ be fixed. For each $\alpha < \xi$, define $S_\alpha(n) = \{ x \in X \ : \ B(x, 1/n) \subseteq V_\alpha \}$. Then define $T_\alpha(n) = S_\alpha(n) \setminus \bigcup_{\beta < \alpha} V_\beta$. This way, $\{T_\alpha(n)\}$ is defined to not only be pairwise disjoint, but so that $d(T_\alpha(n), T_\beta(n)) \ge 1/n$ for $\alpha \ne \beta$: If $x \in T_\alpha(n)$ and $y \in T_\beta(n)$, with $\beta < \alpha$, then $B(y, 1/n) \subseteq V_\beta$ and $x \notin V_\alpha$, so $d(x, y) \ge 1/n$.

So let $E_\alpha(n) = \bigcup_{x \in T_\alpha(n)} B(x, 1/3n)$. By definition, $d(E_\alpha(n), E_\beta(n)) \ge 1/3n$. To see why, if $y_1 \in E_\alpha(n)$ and $y_2 \in E_\beta(n)$ with $\beta < \alpha$, then there exist $x_1 \in T_\alpha(n)$ and $x_2 \in T_\beta(n)$ such that $d(y_1, x_1) < 1/3n$ and $d(y_2, x_2) < 1/3n$. By above, $d(x_1, x_2) \ge 1/n$, so we have that $d(y_1, y_2) \ge 1/3n$. By construction, noting the definition of $S_\alpha(n)$, we see that $E_\alpha(n) \subseteq V_\alpha$, so $\mathcal{E}_n = \{E_\alpha(n)\}$ is an open refinement of $\mathcal{U}$. It is locally finite because for any $x \in X$, the open ball $B(x, 1/6n)$ can only intersect at most one $E_\alpha(n)$, since any other element has distance at least $1/3n$.

Finally, take $\mathcal{E} = \bigcup_{n < \omega} \mathcal{E}_n$. By definition it is a $\sigma$-locally finite refinement of open sets, so we just have to prove it is also a cover. For $x \in X$, let $\alpha$ be the minimum ordinal such that $x \in V_\alpha$. As it is open, let $n \in \N$ be big enough so that $B(x, 1/n) \subseteq V_\alpha$. Then $x \in E_\alpha(n)$. $\square$

For each $n \in \N$, let $\mathcal{B}_n = \{ B(x, 1/n) \ : \ x \in X \}$ be the cover of $1/n$-balls, and let $\mathcal{R}_n$ be the $\sigma$-locally finite refinement which is also a cover, by above. Then $\mathcal{R} = \bigcup_{n < \omega} \mathcal{R}_n$ is also $\sigma$-locally finite, and clearly a basis: If $x \in B(x, 1/n)$, then $x \in \mathcal{B}_n$ and so there is some element $R \in \mathcal{R}_n$ such that $x \in R \subseteq B(x, 1/n)$.

Source: Munkres, Topology: Lemma 39.2 (p. 246) + Theorem 40.3 at step 2 (p. 251)

This is, in essence, one of the directions of the Nagata-Smirnov theorem, which in its usual form, states

$$\text{Metrizable $\iff$ ($T_3$ ∧ Has a $\sigma$-locally finite basis)}$$

Note that if $Y$ is the Kolmogorov quotient of $X$, then $Y$ is $T_3$ and the $\sigma$-locally finite basis of $X$ induces one on $Y$ via the surjective open map $p : X \to Y$. And if $Y$ is metrizable, its metric induces a pseudometric on $X$. So it suffices to prove the $\impliedby$ direction of the metrization theorem above and assume the space is $T_3$.

Lemma 1. (Regular ∧ Has a $\sigma$-locally finite basis) $\implies$ (Normal ∧ $G_\delta$ space)

Proof. Let $\mathcal{B} = \bigcup_{n < \omega} \mathcal{B}_n$ be the basis where each $\mathcal{B}_n$ is locally finite. We use the following facts about locally finite families:

1. If $\mathcal{A}$ is locally finite, then $\mathcal{A}' \subseteq \mathcal{A}$ is locally finite.

2. If $\mathcal{A}$ is locally finite, $\cl{\bigcup_{A \in \mathcal{A}} A} = \bigcup_{A \in \mathcal{A}} \cl{A}$.

(1) is trivial and for (2), $\supseteq$ always holds in general. To prove $\subseteq$, if $x$ is such that every nbd $V$ of $x$ intersects $\bigcup \mathcal{A}$, note that for any fixed nbd $U$, it intersects only finitely many $\{A_1, \dots, A_n\} \subseteq \mathcal{A}$. So because any $U \cap V$ intersects at most those sets, $x \in \cl{\bigcup_{i=1}^n A_i} = \bigcup_{i=1}^n \cl{A_i} \subseteq \bigcup_{A \in \mathcal{A}} \cl{A}$.

To prove $G_\delta$, we’ll prove something a little stronger: Every open set is a countable union of closures of open sets (so in particular, it is $F_\sigma$). Let $V$ be an open set. Define $\mathcal{C}_n = \{ B \in \mathcal{B}_n \ : \ \cl{B} \subseteq V \}$. $\mathcal{C}_n$ is locally finite by (1). If we define $U_n = \bigcup_{B \in \mathcal{C}_n} B$, then by (2), $\cl{U_n} = \bigcup_{B \in \mathcal{C}_n} \cl{B} \subseteq V$. We wish to prove $V = \bigcup_{n < \omega} \cl{U_n}$, and we’ve just proven $\supseteq$. For $\subseteq$, let $x \in V$. As the space is regular, there is an open nbd $U$ of $x$ such that $x \in \cl{U} \subseteq V$. $\mathcal{B}$ is a basis, so there is some $B \in \mathcal{B}_n$ such that $x \in B \subseteq U$, and so $\cl{B} \subseteq V$ implies $x \in \cl{U_n}$.

To prove normality, let $A,B$ be closed disjoint. By above, let $\{V_n\}$ be open sets with $X \setminus A = \bigcup_{n < \omega} \cl{V_n}$ and $\{U_n\}$ be open sets with $X \setminus B = \bigcup_{n < \omega} \cl{U_n}$. We now construct the separation in a similar fashion to (T26). Define

$$U = \bigcup_{n < \omega} \left(U_n \setminus \bigcup_{k \le n} \cl{V_k}\right) \qquad V = \bigcup_{n < \omega} \left(V_n \setminus \bigcup_{k \le n} \cl{U_k}\right)$$

Then $A \subseteq U$, $B \subseteq V$, and $U \cap V = \emptyset$. $\square$

Lemma 2. If $X$ is $T_1$ and If $f : X \to [0, 1]^I$ is a map such that

1. $f_\alpha : X \to [0, 1]$ is continuous for each $\alpha \in I$.

2. For every $x_0$ and nbd $U$ of $x_0$, there is some $\alpha \in I$ such that $f_\alpha(x_0) > 0$ and $f_\alpha(X \setminus U) = \{0\}$.

Then $f$ is an embedding (a homeomorphism to its image).

Proof. As we’re taking $[0, 1]^I$ with the product topology, (1) implies $f$ is continuous. Since $X$ is $T_1$, for $x \ne y$, let $x \in U$ with $y \notin U$ and so for some $\alpha$, $f_\alpha(x) > 0$ while $f_\alpha(y) = 0$ by (2). Now let $V$ be open in $X$. For each $f(x) = y \in f(V)$, let $\alpha$ such that $f_\alpha(x_0) > 0$ and $f_\alpha(X \setminus V) = \{0\}$. Then $W = \pi_\alpha^{-1}((0, \infty))$ is a nbd of $f(x)$, so that $W \cap f(X) \subseteq f(V)$. This proves $f$ is open. $\square$

Now we’re ready to prove: ($T_3$ ∧ Has a $\sigma$-locally finite basis) $\implies$ Metrizable.

Let $\mathcal{B} = \bigcup_{n < \omega} \mathcal{B}_n$ be the basis where each $\mathcal{B}_n$ is locally finite. Let $I = \{ (n, B) \in \N \times \mathcal{B} \ : \ B \in \mathcal{B}_n \}$. By Lemma 1, and (T257), we see that each $X \setminus B$ is a cozero set, so there exists a continuous map $f : X \to [0, 1]$ with $X \setminus B = f^{-1}(0)$. Using the axiom of choice, for each $(n, B) \in I$, choose a function $f_{(n, B)} : X \to [0, 1/n]$ such that $f_{(n, B)}^{-1}(0) = B$. Then clearly the map $f : X \to [0, 1]^I$ with $f_{(n, B)}$ as each coordinate satisfies the statements in Lemma 2, since for $x_0 \in U$, there is some $B \in \mathcal{B}_n$ with $x_0 \in B \subseteq U$, and $g = f_{(n, B)}$ satisfies $g(x_0) = 0$ and $g(X \setminus U) > 0$.

However, $f : X \to [0, 1]^I$ is only a homeomorphism to $f(X)$ as a subspace of the product topology on $[0, 1]^I$. We need to show it is as well as a subspace of the uniform topology on $[0, 1]^I$, which is the one induced by the metric $\rho(x, y) = \sup_{i \in I} \abs{x_i - y_i}$. We already know it is injective, and if $V$ is open, $f(V)$ is open in the product topology, so it is also open in the metric topology as it is finer. We merely have to show $f$ is still continuous.

Let $x_0 \in X$ and $\varepsilon > 0$. Choose $n_0$ big enough so that $1/n_0 \le \varepsilon$. For $n \ge n_0$, the range of $f_{(n, B)}$ lies in $[0, 1/n]$, so $\abs{f_{(n, B)}(x) - f_{(n, B)}(x_0)} < \varepsilon$ is immediate. For each $n < n_0$, we know some nbd $U(n)$ of $x_0$ intersects finitely many $\{B_1, \dots, B_m\} \subseteq \mathcal{B}_n$. This means $f_{(n, B)}(U) = \{0\}$ for every other $B \in \mathcal{B}_n$, yet $f_{(n, B_i)}(x_0) > 0$ for each $1 \le i \le m$. By continuity, choose a $V_i(n) \subseteq U(n)$ nbd of $x_0$ such that for all $x \in V_i(n)$, $\abs{f_{(n, B_i)}(x_0) - f_{(n, B_i)}(x)} < \varepsilon$. Then define $V(n) = \bigcup_{i=1}^m V_i(n)$ and $V = \bigcap_{n < n_0} V(n)$, which is open, and has the property that for any $x \in V$, we have as we wished, $\rho(f(x), f(x_0)) < \varepsilon$.

Source: Munkres, Topology: Theorems 39.1 (for Lemma 1), 34.2 (for Lemma 2), 40.3 (main result)