Sequentially compact metric space is compact

I'm proving this important result. Could you have a check on my proof?

Let $(X, d)$ be sequentially compact metric space. Then $X$ is compact.

We need $2$ lemmas.

• Lemma 1: Sequentially compact metric space is totally bounded. [A proof is given here]

• Lemma 2: A metric space is Lindelöf if and only if it's separable. [A proof is given here]

By lemma 1, $X$ is totally bounded. Then for each $n \in \mathbb N^*$, there is a finite cover $C_n$ consisting of open balls with radius $1/n$. Let $C$ be the set of centers of balls from those $C_n$. Then $C$ is a countable dense subset of $X$, so $X$ is separable. By lemma 2, $X$ is Lindelöf. Let $(O_i)_{i\in I}$ be an open cover of $X$. Then there is a countable subset $J$ of $I$ such that $(O_i)_{i\in J}$ is a countable cover.

WLOG, we relabel and assume $J:=\mathbb N$. Assume that $(O_n)_{n\in \mathbb N}$ has no finite subcover. Let $F_n:= \bigcap_{k \le n} O^c_k$ Then $(F_n)$ is a decreasing sequence of closed sets. Then we pick $x_0 \in O_0$ and $x_{n+1} \in O_{n+1} \cap F_n$. Because $X$ is sequentially compact, there exist $a\in X$ and a subsequence $(x_{n_m})$ such that $x_{n_m} \to a$ as $m \to \infty$. We have $x_{n_m} \in F_n$ for all $n_m > n$, so $a \in F_n$ for all $n$. Hence $\bigcap_n F_n \neq \emptyset$, which is a contradiction. As such, $(O_n)_{n\in \mathbb N}$ has a finite subcover.

Solution 1:

Easier: if $(X,d)$ is sequentially compact $X$ does not contain an uncountable discrete closed set so by the result I showed here, $X$ is Lindelöf.

Also (quite generally) $X$ is countably compact (this implication "sequentially compact implies countably compact" holds in any space).

A Lindelöf countably compact space is trivially compact.

For the reverse, note that a compact space is countably compact and hence sequentially compact as $(X,d)$ is first countable hence sequential (see here).