If not discrete, there exists a non-isolated point $x$. Let $V$ be its injectively path connected nbd, which must contain some $y \ne x$. Let $f : [0, 1] \to X$ be an injective path from $x$ to $y$. Then $f([0, 1/2))$ and $f((1/2, 1])$ are connected and disjoint.

An arc is, by definition, an embedding (hence, injective).

An arc is, by definition, an embedding (hence, injective).

Every $p \in \emptyset$ is homeomorphic to $\R$.

The complement of any set must be finite.

If $A$ is infinite, let $B \subseteq A$ countably infinite, which must be discrete, so $\{x\}$ for $x \in B$ is a countable cover with no finite subcover. Thus, countably compact sets must be finite, which are closed, as $X$ is $T_1$ (T221).