Show that the closure of a subset is bounded if the subset is bounded

If $A$ is bounded then there exists some closed ball $B$ such that $A \subset B$.

Now by definition of $\bar A$ as the smallest closed set that contains $A$, it must be that $\bar A \subset B$.

This proves that $\bar A$ is bounded.

No computation involved in this proof !

Either definition works:

If $a$ is a limit point of $A$, there is some sequence $a_n \in A $such that $a_n \rightarrow a$. It then follows that $d(a,x_0) \leq d(a,a_n) + d(a_n,x_0) \leq d(a,a_n)+K$ for any $n$. Letting $n \rightarrow \infty$ we get that $d(a_n,a) \rightarrow 0$ so $d(a,x_0) \leq K$. (Since the left hand side is independent of $n$.)

Otherwise, for any $a \in \bar A$, $B_1(a)\cap A \not = \emptyset$, so we can pick a point $a'$ in the intersection. Hence $d(a,x_0) \leq d(a,a') + d(a',x_0) \leq 1 + K$. (You can even show that $\bar A$ is bounded by the same $K$ as before by noting that the above is true with $1$ replaced by any $\epsilon >0.$)

Let $y\in\mathrm{cl}A$, then $y=\lim\limits_{n\to\infty}x_n$ for some $\{x_n:n\in\mathbb{N}\}\subset A$, since $d:X\times X\to\mathbb{R}_+$ is continuous then $$ d(y,x_0)=d(\lim\limits_{n\to\infty} x_n,x_0)=\lim\limits_{n\to\infty}d(x_n,x_0)\leq K $$