Is this proof about boundedness of Cauchy sequences correct? A school exercise reads as follows:
"Show that every Cauchy sequence $(x_n)_{x \in \mathbb{N}}$ in a metric space $(X, d)$ is bounded"
Here is the solution(proof) provided by the lecturer:
"
We show that there is some $r > 0$ and some $x \in X$ such that $\{x_n\} \subseteq B_r(x)$. Since
$\{x_n\}_n$ is Cauchy, there is some $N \in \mathbb{N}$ such that for all $n, m \geq N$
$$
d(x_n, x_m) < 1
$$
Let
$$
s := \underset{1 \leq i \leq N}{\text{max}} d(x_1, x_i)
$$
Then if $n \leq N$
$$
d(x_n, x_1) \leq s,
$$
while if $n \geq N$, by the triangle inequality,
$$
d(x_n, x_1) \leq d(x_1, x_N) + d(x_N, x_n) < s + 1
$$
It follows that for $r = s+1$, we have $\{x_n\} \subseteq B_r(x_1)$.
"
Here is my questioning of the validity of this proof:
My issue is with this particular step

Let
$$
s := \underset{1 \leq i \leq N}{\text{max}} d(x_1, x_i)
$$

It isn't obvious to me that this maximum exists. It seems to me that making the assumption that the maximum exists is exactly the same as making the assumption that $\{x_n\}_{x \in \mathbb{N}}$ is bounded. Basically, it seems to me that the proof says the following: "If you assume that $\{x_n\}_{x \in \mathbb{N}}$ is bounded, then $\{x_n\}_{x \in \mathbb{N}}$ is bounded, i.e circular reasoning. Am I wrong? If so, can someone prove to me why the maximum has to exist?

The maximum is over finitely many real numbers, and so of course it exists. You are right that it would be circular to take the maximum over all $(i, j)$ pairs, but your professor is taking the maximum only over $N^2$ pairs. This is a fairly common technique in analysis: Somehow reduce an infinitary question to a finitary question.
