If $X$ is compact metric space, then $X$ is bounded

Assume that $X$ is metric space and compact. We must proof that $X$ is bounded.

Now, at first we notice that $X$ is sequentially compact. But then what?

We must find some $W\geq 0$ such that $d(x,y)< W$ by all $x,y\in X$.

How I can do that? Need a hint.

Pick $p \in X$. Then $O_n = B(p,n)$, $n \in \mathbb{N}$ is a cover of $X$. There is a finite subcover of the $O_n$, and as $O_1 \subset O_2 \subset O_3 \ldots$, already one of them (the largest index) already covers $X$ and so $X$ is bounded.
Hint: Fix some $x\in X$. For $n\in \Bbb{N}$, let $B_n=\{y \in X \mid d(x,y)<n\}$ be the open ball. What is $\bigcup_{n \in \Bbb{N}}B_n$? What does the compacity of $X$ tell you about this union?
Hint: (If you have in hand the theorem stating that a continuous real-valued function on a compact space is bounded.) Consider the function $f:X\to[0,\infty)$ defined by $f(x) =d(x,x_0)$, where $x_0$ is a fixed point of $X$. Show that $f$ is continuous, hence bounded, and then use $d(x,y)\le d(x,x_0)+d(y,x_0)$.
• I would have done similarly, but have noted that $X\times X$ is compact as well, so that $d(X\times X)$, which is the set of all distances between two points of $X$, is a compact subset of $\Bbb R^{\ge0}$, and thus bounded. Oct 30 '15 at 22:42