Compact Subspace of Metric Space is Bounded

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Let $$C$$ be a subspace of $$M$$.

If $$C$$ is compact, then it is bounded.

Proof
Let $$a \in M$$.

Let $$N_n \left({a}\right)$$ be the $n$-neighborhood of $$a$$.

Then $$C \subseteq \bigcup_{n=1}^\infty N_n \left({a}\right)$$ because $$\forall x \in C: d \left({x, a}\right) < n$$ for some $$n \in \N$$.

Thus the collection $$\left\{{N_n \left({a}\right): n \in \N}\right\}$$ forms an open cover of $$C$$.

Because $$C$$ is compact, it has a finite subcover, say: $$\left\{{N_{n_1} \left({a}\right), N_{n_2} \left({a}\right), \ldots, N_{n_r} \left({a}\right)}\right\}$$.

Let $$n = \max \left\{{n_1, n_2, \ldots, n_r}\right\}$$.

Then $$C \subseteq \bigcup_{n=1}^r N_{n_r} \left({a}\right) = N_n \left({a}\right)$$.

The result follows by definition of bounded.