Sequentially Compact Metric Space is Totally Bounded/Proof 2

Lemma
A sequentially compact metric space $\left({X, d}\right)$ is totally bounded.

Proof
Let us reason by contradiction.

If the result is false, then there is some $\epsilon > 0$ for which it is impossible to cover $X$ with a finite number of balls of radius $\epsilon$.

Let us construct an infinite sequence $x_1, x_2,\ldots$ recursively as follows:
 * We choose any $x_1 \in X$.
 * Once we have chosen $x_1,\ldots,x_n$ (for some $n \geq 1$), choose $x_{n+1}$ such that
 * $\displaystyle x_{n+1} \notin \bigcup_{i=1}^n B(x_i,\epsilon)$

One can always choose $x_{n+1}$ with the above property because the finite collection of balls $B(x_1,\epsilon),\ldots,B(x_n,\epsilon)$ cannot cover $X$ due to our initial assumption.

So, this process constructs an infinite sequence $\{x_i\}_{i \in \N}$.

Now, as $X$ is sequentially compact, the sequence $\{x_i\}$ has a subsequence which converges to some point $x \in X$.

We will prove next that this is a contradiction, as the points $x_i$ are more than a fixed distance apart.

As a subsequence of $\{x_i\}$ converges to $x$, one can find two members $x_N$, $x_M$ of the sequence such that:
 * $d(x_N, x) < \epsilon/4 \quad \text{ and } \quad d(x_M, x) < \epsilon/4$.

(Actually, we can find an infinite number of terms with this property in the sequence, but two will do).

We name $x_N, x_M$ such that $N > M$.

Then:
 * $d(x_N, x_M) \leq d(x_N, x) + d(x, x_M) < \epsilon/4 + \epsilon/4 = \epsilon/2$.

This is a contradiction, as $x_M \notin B(x_N, \epsilon)$ by the way we constructed the sequence.