In proving the equivalance of sequential compactness and compactness in metric spaces, I come across this lemma. Could you verify if it's fine?
Let $(X, d)$ be a sequentially compact metric space. Then $X$ is complete and totally bounded.
My attempt: Let $(x_n)$ be a Cauchy sequence in $X$. Because $X$ is sequentially compact, there exist $a \in X$ and a subsequence $(x_{n_k})$ such that $x_{n_k} \to a$ as $k \to \infty$. Fix $\varepsilon > 0$, there is $N$ s.t. $d(x_n,x_m) < \varepsilon$ for all $n,m \ge N$. Also, there is $N'$ s.t. $d(x_{n_k}, a) < \varepsilon$ for all $k \ge N'$. Let $N'' := \max\{N, N'\}$. Then $d(x_n, a) \le d(x_n, x_{n_{N''}}) + d(x_{n_{N''}}, a)<2\varepsilon$ for all $n \ge N$. Hence $x_n \to a$ and thus $X$ is complete.
Assume the contrary that $X$ is not totally bounded. Then there is $r >0$ s.t. no finite collection of open balls with radius $r$ covers $X$. We pick some $x_0 \in X$ and define a sequence $(x_n)$ recursively by $$x_{n+1} \notin \bigcup_{k \le n} \mathbb B(x_n, r).$$
This implies $$d(x_{n}, x_m) \ge r, \quad \forall n \neq m.$$
Because $X$ is sequentially compact, there exist $a \in X$ and a subsequence $(x_{n_k})$ such that $x_{n_k} \to a$ as $k \to \infty$. Clearly, $(x_{n_k})$ is a Cauchy sequence. This means there is $N$ such that $d(x_{n_k}, x_{n_h}) < r$ for all $k,h \ge N$. However, this contradicts the fact that $d(x_{n_k}, x_{n_h}) \ge r$ for all $k \neq h$. Hence $X$ is totally bounded.
