Sequentially Compact Metric Space is Totally Bounded/Proof 2

The following result states that a sequentially compact metric space is totally bounded. There is a stronger result than this, namely, that a sequentially compact metric space is actually compact; this is stronger because being compact is equivalent to being totally bounded and complete. The lemma below is one of the steps of a possible proof of the stronger result we just mentioned, and hence we state it as a lemma, and not as a theorem.

Lemma
A sequentially compact metric space $$(X,d)$$ 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 $$\epsilon$$.

Ley 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
 * $$x_{n+1} \notin \bigcup_{i=1}^n B(x_i,\epsilon)$$.

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
 * $$\vert x_N - x \vert < \epsilon/4 \quad \text{ and } \vert x_M - x \vert < \epsilon/4.$$

(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
 * $$\vert x_N - x_M \vert \leq \vert x_N - x \vert + \vert x - x_M \vert < \epsilon/4 + \epsilon/4 = \epsilon/2.$$

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