It suffices to show the space is normal. We first show regularity. Let $A$ be closed and $x \notin A$. Let $K$ be a compact nbd of $x$. $A \cap K$ is compact, and $T_2$ spaces separate points from compact sets. So let $x \in U$ and $A \cap K \subseteq V$ with $U \cap V = \emptyset$. Choose $W = U \cap \text{int}(K)$ and $W' = V \cup (X \setminus K)$. Then $W \cap W' = \emptyset$ are open with $x \in W$ and $A \subseteq W'$.

From this point, since the space is $\sigma$-compact, it’s possible to use other theorems of $\pi$-base to prove normality: click here. It feels a bit like cheating, so we instead provide a more direct approach.

Let $A$ and $B$ be closed sets. By regularity, for each $x \in A$, there are nbds $x \in U_x$ and $B \subseteq V_x$ which are disjoint. Thus, $\cl{U_x} \cap B = \emptyset$. If we define $\mathcal{U}_x$ as the nbds of $x$ with this property, then it’s nonempty and $\mathcal{U} = \bigcup_{x \in X} \mathcal{U}_x$ is an open cover of $A$. $A$ is closed and $X$ is $\sigma$-compact, which clearly implies it is Lindelöf (T122), so $A$ is Lindelöf and we can extract a countable subcover $\{U_n\}$. We can do the exact same to $B$, finding a countable open cover $\{V_n\}$ such that $\cl{V_n} \cap A = \emptyset$ for each $n$. We then define

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

$U$ and $V$ are open by construction. For each $x \in A$, $x \in U_n$ for some $n$ and since $\cl{V_k} \cap A = \emptyset$, we have $x \in U$. So $A \subseteq U$ and through a similar argument, $B \subseteq V$. Now we show $U \cap V = \emptyset$. Assume $x \in U \cap V$, which means $x \in U_n \setminus \bigcup_{k \le n} \cl{V_k}$ and $x \in V_m \setminus \bigcup_{k \le m} \cl{U_k}$. If $n \le m$, then $x \in U_n$ and $x \notin \cl{U_n}$ generates a contradiction, and $n \ge m$ would imply $x \in V_m$ yet $x \notin \cl{V_m}$.

Denote $x \sim y$ to mean $x,y$ are indistinguishable. Let $Y = X/{\sim}$ be the Kolmogorov quotient and $q : X \to Y$ the surjective map. Note that it is continuous and an open map, therefore a closed map, as it is surjective. This means two things.

1. If $x \notin A$ with $A \subseteq X$ closed, then $q(x) \notin q(A)$ with $q(A)$ closed, and a function $f : Y \to [0, 1]$ separating $q(x)$ and $q(A)$ induces a continuous function $g : X \to [0, 1]$ separating $x$ and $A$. This proves $Y$ completely regular $\implies X$ completely regular.

2. If $x \in V \subseteq K$ with $V$ open and $K$ compact, then $q(x) \in q(V) \subseteq q(K)$, where $q(V)$ is open as $q$ is open, and $q(K)$ compact because $q$ is continuous. So $X$ weakly locally compact $\implies Y$ weakly locally compact.

Since $X$ is $R_1$, then $Y$ is $T_2$. Joining facts (1) and (2), it suffices to prove (Weakly locally compact ∧ $T_2$) $\implies$ $T_{3 \frac 1 2}$.

We first prove locally compactness. To do that, we’ll prove regularity and then then the result follows from (T246). Let $x \notin A$ with $A$ closed. Let $x \in V \subseteq K$ with $V$ open and $K$ compact, by weakly locally compact. Note that $T_2$ spaces separate points from compact sets, so let $x \in U$ and $A \cap K \subseteq U'$ with $U \cap U' = \emptyset$. As $K$ is closed, we see that $A \subseteq U' \cup (X \setminus K)$, and $(U \cap V) \cap (U' \cup (X \setminus K)) = \emptyset$, which separates $x$ from $A$.

Since $Y$ is locally compact and $T_2$, its one-point compactification $Y^* = Y \cup \{\infty\}$ is compact $T_2$, which must be $T_4$, since $T_2$ spaces separate compact sets and any closed sets of $Y^*$ are compact. Normal spaces are completely regular (T37), so $Y^*$ is completely regular. Let $x \in Y$ and $A \subseteq Y$ closed. Then $A$ is also closed in $Y^*$, so let $f : Y^* \to [0, 1]$ with $f(A) = \{0\}$ and $f(x) = 1$, and $f|_Y : Y \to [0, 1]$ is still continuous with the same properties.

Let $(x_n)$ be a sequence. By contradiction, suppose it has no converging subsequence and for each $k < \omega$, define $A[k] = \bigcup_{m \ge k} \cl{\{x_m\}}$. We first prove each $A[k]$ is closed. As the space is sequential, it suffices to take $(y_n)$ a sequence in $A[k]$ with $y_n \to y$ and prove $y \in A[k]$. Let $Y(m) = \{ n > m \ : \ y_n \in \cl{\{x_m\}}\}$ for each $m \ge k$. If some $Y(m) is infinite, we could construct a subsequence $(y_{f(n)})$ in $\cl{\{x_m\}}$ such that $y_{f(n)} \to y$, and so $y \in \cl{\{x_m\}} \subseteq A[k]$. Otherwise, each $Y(m)$ is finite and infinitely many are nonempty. So let $m_1$ be the least $m$ for which $Y(m) \ne \emptyset$ and $n_1 \in Y(m_1)$. Inductively, let $m_{j+1}$ be the least $m > n_j$ such that $Y(m) \ne \emptyset$ and $n_{j+1} \in Y(m_{j+1})$. Now we’ve chosen subsequences $(y_{n_j})$ and $(x_{m_j})$ such that $y_{n_j} \in \cl{\{x_{m_j}\}}$. For any $V$ nbd of $y$, there exists some $n_0$ such that $y_{n_j} \in V$ for $j \ge n_0$. But then $V$ is a nbd of $y_{n_j}$, so $x_{m_j} \in V$ for all $j \ge n_0$. As this holds for all $V$, that would imply $x_{n_j} \to y$, a contradiction!

So each $A[k]$ is closed, and as the space is countably compact, let $p$ be an accumulation point of $(x_n)$. Then clearly $p$ is a limit point of $A[k]$ for all $k$. Now consider the sets $X(n) = \{ m > n \ : \ p \in \cl{\{x_m\}} \}$. If $X(k) = \emptyset$ for some $k$, that would imply $p \notin A[k+1]$, which is impossible since $p \in \cl{A[k+1]}$ and $A[k+1]$ is closed. So let $m_1 = \min X(1)$ and $m_{k+1} = \min X(m_k)$. We’ve contructed a sequence $(x_{m_k})$ such that $p \in \cl{\{x_{m_k}\}}$ for all $k$. Then for any $V$ nbd of $p$, $x_{m_k} \in V$ and that would imply $x_{m_k} \to p$.

Source: Sequential space methods, Kremsater

We first prove that for every decreasing sequence $\{F_n\}$ of closed sets with $\bigcap_{n < \omega} F_n = \emptyset$, there exists a sequence $\{A_n\}$ of closed $G_\delta$-sets such that $\bigcap_{n < \omega} A_n = \emptyset$ and $F_n \subseteq A_n$ for all $n$. Let $\{F_n\}$ be such a sequence. Using Ishikawa’s characterization, there exists a sequence of open sets $\{U_n\}$ with $F_n \subseteq U_n$ and $\bigcap_{n < \omega} U_n = \emptyset$. That means $F_n \cap F'_n = \emptyset$ where $F'_n = U_n^C$. By Urysohn’s lemma, let $f_n : X \to [0, 1]$ with $f(F_n) \subseteq \{0\}$ and $f(F'_n) \subseteq \{1\}$. Then define $V_{nk} = f_n^{-1}([0, 1/k))$, which proves $A_n = \bigcap_{k=1}^\infty V_{nk} = f^{-1}(0)$ is a closed $G_\delta$-set with $F_n \subseteq A_n$. Now to show $\{A_n\}$ has empty intersection, we just have to show $A_n \subseteq U_n$ for each $n$. But $x \notin U_n \implies f(x) = 1 \implies f(x) \ne 0 \implies x \notin A_n$.

Now let $\{U_n\}$ be a countable open cover. Define $F_n = X \setminus \bigcup_{k \le n} U_n$. Then $\{F_n\}$ is decreasing and must have empty intersection. Let $\{A_n\}$ be the closed $G_\delta$-sets with empty intersection such that $F_n \subseteq A_n$. Then each $B_n = X \setminus A_n$ is an $F_\sigma$-set, so we write $B_n = \bigcup_{k < \omega} F'_{nk}$. Note that by the way we constructed the $V_{nk}$ above, we could ensure that $V_{nk} \supseteq \cl{V_{n(k+1)}}$. So here, we can assume $\{F'_{nk}\}_{k < \omega}$ is increasing with $F'_{nk} \subseteq \text{int}(F'_{n(k+1)})$. Define $H_{nk} = \text{int}(F'_{nk})$. Then $B_n = \bigcup_{k < \omega} H_{nk}$ and $F'_{nk} \subseteq B_n \subseteq \bigcup_{k \le n} U_n$.

If we define $V_n = U_n \setminus \bigcup_{k < n} F'_{nk}$, each $V_n$ is open. For each $x$, let $n$ be the first index with $x \in U_n$ and so $x \notin \bigcup_{k < n} U_k \implies x \notin F'_{nk}$ for each $k < n$, so that $x \in V_n$, proving $\{V_n\}$ is an open refinement of $\{U_n\}$. Finally, since $\bigcap_{n < \omega} A_n = \emptyset$, there must be some $n$ with $x \notin A_n$, meaning $x \in F'_{n(k-1)} \subseteq H_{nk}$ for some $k$. Then for $i > \max \{n, k\}$, $H_{nk} \subseteq F'_{ni}$ so that $H_{nk} \cap V_i = \emptyset$. So $H_{nk}$ is a nbd of $x$ which intersects at most $V_n$ with $n < i$, proving it is locally finite.

Source: C. H. Dowker, On Countably Paracompact Spaces, Theorem 2 (d) ⇒ (a)