Compact First-Countable Space is Sequentially Compact

Theorem
Let $T = \left({X, \vartheta}\right)$ be a first-countable topological space which is compact.

Then every sequence in $X$ has a convergent subsequence, i.e. $T$ is sequentially compact.

Proof
We prove the contrapositive.

Suppose the sequence $\left \langle {x_n} \right \rangle_{n \in \N}$ in $X$ has no convergent subsequence. That is, for every point $x \in X$, there is no subsequence of $\left \langle {x_n} \right \rangle$ that converges to $x$.

We first argue by contradiction that we can select an open set around each point in $X$ which contains only finitely many of the terms in the sequence:

Let $x\in X$ and suppose every open set containing $x$ also contains infinitely many terms of the sequence $\left \langle {x_n} \right \rangle$.


 * Since $X$ is first-countable, there is a countable local basis $\{U'_0,U'_1, \dots\}$ at $x$. Since finite intersections of open sets are open and are subsets of those terms of the intersection, we can construct another countable local basis $\{U_n\}$ at $x$ by defining $U_n$ to be $U'_0 \cap U'_1 \cap \cdots \cap U'_n$.  This gives us a countable local basis at $x$ such that $U_{n+1} \subseteq U_n$ for all $n\in\mathbb{N}$.


 * Now, since each $U_n$ has infinitely many terms from $\left \langle {x_n} \right \rangle$ by our assumption, we can construct a subsequence $\left \langle {x_{n_k}} \right \rangle$ such that $x_{n_k}\in U_k$ as follows:


 * Let $x_{n_0}$ be any term in the sequence which is contained in $U_0$.


 * When $x_{n_k}$ has been defined, let $x_{n_{k+1}}$ be one of the infinitely many terms in the sequence which is contained in $U_k$ and whose index is larger than $n_k$.


 * We now show that this sequence converges to $x$:


 * Let $U$ be an open set containing $x$.


 * $U$ contains some $U_K$ as a subset since $\{U_k : k\in \mathbb N\}$ is a local basis.


 * $U_K$ is a subset of each $U_{k}$ for $k > K$.


 * But, we constructed $\left \langle {x_{n_k}} \right \rangle$ so that $x_{n_k} \in U_k$


 * Hence, for all $k > K$, $x_{n_k} \in U_k \subseteq U_K \subseteq U$.


 * This contradicts the assumption that $\left \langle {x_n} \right \rangle$ has no convergent subsequence.

Thus, for each $x\in X$, we can select an open set $U_x$ such that $x\in U_x$ and $U_x$ contains only finitely many terms from the sequence $\left \langle {x_n} \right \rangle$.

The family $\mathcal U = \left\{{U_x : x \in X}\right\}$ is clearly an open cover of $X$.

Now, suppose $U$ has a finite subcover $\{U_{x_1}, U_{x_2}, \dots, U_{x_m}\}$.


 * Then $U_{x_1} \cup U_{x_2} \cup \cdots \cup U_{x_m}$ contains all of $X$ since it is a cover, but contains only finitely many terms of the sequence $\left \langle {x_n} \right \rangle$ since it is a finite union of sets which each contain only finitely many such terms.


 * However, this is impossible, since $\left \langle {x_n} \right \rangle$ is a sequence in $X$, and all of its infinitely many terms are in $X$.

Thus, we have found an open cover of $X$ which has no finite subcover.

By definition, this means that $X$ is not compact.

The theorem follows by contraposition.