Compact First-Countable Space is Sequentially Compact
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Theorem
Let $T = \left({S, \tau}\right)$ be a compact first-countable topological space.
Then every infinite sequence in $S$ has a convergent subsequence; i.e., $T$ is sequentially compact.
Proof
Let $\left \langle {x_n} \right \rangle_{n \mathop \in \N}$ be an infinite sequence in $S$.
Aiming for a contradiction, suppose that $\left \langle {x_n} \right \rangle_{n \mathop \in \N}$ does not have a convergent subsequence.
By Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit, it follows that $\left \langle {x_n} \right \rangle_{n \mathop \in \N}$ does not have an accumulation point in $S$.
Thus, for each $x \in S$, we can select an open set $U_x$ such that $x \in U_x$ and $U_x$ only contains $x_n$ for finitely many $n \in \N$.
The set $\mathcal U = \left\{{U_x : x \in X}\right\}$ is clearly an open cover for $S$.
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By the definition of a compact space, there exists a finite subcover $\{{U_{x_1}, U_{x_2}, \ldots, U_{x_m}}\}$ of $\mathcal U$ for $X$.
Then $U_{x_1} \cup U_{x_2} \cup \cdots \cup U_{x_m}$ contains all of $S$ since it is a cover.
However, each open set in this union only contains $x_n$ for a finite number of $n$, and this is a finite union of sets. So, the union only contains $x_n$ for finitely many $n$.
This implies that $x_n \in X$ for only finitely many $n \in \N$.
This is a contradiction, since $\left \langle {x_n} \right \rangle$ is a sequence in $S$, and so $x_n \in S$ for all $n \in \N$.
$\blacksquare$
To prove this in ZF, one could simply use Compact Space is Countably Compact and Countably Compact First-Countable Space is Sequentially Compact.
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