TITLE: Proving that sequentially compact spaces are compact.
QUESTION [8 upvotes]: I remember seeing this proof somewhere (perhaps here, but I don't remember where) that goes something like this.
Suppose $X$ is sequentially compact, and by contradiction suppose $\{U_n\}$ is a countable open cover with no finite subcover.  Then for any positive integer $n$, the set $\{U_i : i \le n\}$ is not an open cover, so there exists $x_n \notin \bigcup_{i \le n} U_i$.  Hence, we obtain sequence, and by sequential compactness, there exists a subsequence $x_{n_j}$ that converges to $a \in X$.  However, $ a \in U_k$ for some positive integer $k$ and by construction, $x_{n_j} \notin U_k$ if $n_j \ge k$. This is a contradiction.
Doesn't this only prove every countable open cover must have a finite subcover?

REPLY [3 votes]: For metric spaces, here is a proof of the equivalence.
Let $ (X,d) $ be a metric space, and $ A \subseteq X $.
Thm: $ A $ is sequentially compact if and only if every open cover of $ A $ has a finite subcover.
Pf: $ \underline{\boldsymbol{\implies}} $ Say $ A $ is sequentially compact, and $ \{ U_i \}_{i\in I} $ is an open cover of $ A $.
Firstly there is a $ \delta > 0 $ such that any ball $ B(x, \delta) $ with $ x \in A $ is contained in some $ U_i $.

Suppose not. Then for every integer $ n > 0 $ there is an $ x_n \in A $ such that $ B(x_n , \frac{1}{n} ) $ isnt fully contained in any $ U_i $. Pass to a subsequence $ (x_j)_{j \in J} $, $ J \subseteq \mathbb{Z}_{>0} $ which converges to a point $ p \in A $ as $ j \in J, j \to \infty$.
Now $ p $ is contained in some $ U_k $, and there is an $ \epsilon > 0 $ such that $ B(p, \epsilon) \subseteq U_k $. Picking a $ j \in J $ such that $ \{ d(x_j, p) < \frac{\epsilon}{2} ; \frac{1}{j} < \frac{\epsilon}{2} \}$, we see $ B(x_j, \frac{1}{j}) \subseteq B(p, \epsilon) \subseteq U_k $, a contradiction.

There are finitely many open balls, with radius $ \delta $ and centres in $ A $, whose union contains $ A $.

[Corrected] Suppose not. Pick $ x_1 \in A $. As $ B(x_1, \delta) $ doesnt cover $ A $, pick $ x_2 \in A \setminus B(x_1, \delta) $. As $ B(x_1, \delta) \cup B(x_2, \delta) $ doesnt cover $ A $, let $ x_3 \in A \setminus (B(x_1, \delta) \cup B(x_2, \delta)) $, and so on. Pass to a subsequence $ (x_j)_{j \in J} $, $ J \subseteq \mathbb{Z}_{>0} $ convergent to a point $ p \in A $ as $ j \in J, j \to \infty $.
But ${ d(x _{j}, x _{j'}) \geq \delta }$ for all distinct ${ j, j' }$ in ${ J }.$ Especially ${ (x _j) _{j \in J} }$ isnt Cauchy, a contradiction.

So there are finitely many balls $ B(x_1, \delta), \ldots, B(x_n, \delta) $ with centres in $ A $, whose union contains $ A $. Also each of these balls is contained in some $ U_i $, giving a finite subcover.
$ \underline{\boldsymbol{\impliedby}}$ Say every open cover of $ A $ has a finite subcover, and let $ (x_n) $ be a sequence in $ A $.
Seq $ (x_n) $ has an accumulation point in $ A $.

Suppose not. So for every $ y \in A $ there is a $ \delta_y > 0 $ such that $ x_n \in B(y, \delta_y) $ for only finitely many $ n $. Now open cover $ \{ B(y, \delta_y) : y \in A \} $ of $ A $ has a finite subcover $ \{ B(y_1, \delta_{y_1}), \ldots, B(y_k, \delta_{y_k}) \} $. But there are only finitely many indices $ n $ such that $ x_n \in B(y_1, \delta_{y_1}) \cup \ldots \cup B(y_k, \delta_{y_k}) $, a contradiction.