Proof:

The student is advised to draw relevant pictures as he reads on. Suppose that a cover $\{G_{\alpha}\}$ has no Lebesgue number. Then for every $n\in \mathbb{N}$, $1/n$ is not a Lebesgue number and so there is a point $x_n \in X$ such that the ball of radius $1/n$ centered at $x_n$ is not contained in any of the open sets in the covering. By compactness the sequence $\{x_n\}$ has a convergent subsequence converging to a point $p \in X$. Choose an $\alpha$ such that $G_{\alpha}$ contains $p$ and there is a $\delta > 0$ such that the ball of radius $\delta$ around $p$ is contained in $G_{\alpha}$. Now take $n$ large enough that $1/n < \delta/3$ and $x_n$ is contained in the ball of radius $\delta/3$ centered at $p$.

Now, since the ball of radius $1/n$ with center $x_n$ is not contained in any of the open sets in our covering, there exists $y_n \in X$ such that $y_n \notin G_{\alpha}$ and $d(x_n, y_n) < 1/n$. But

$$d(p, y_n) \leq d(p, x_n) + d(x_n, y_n) < 2\delta/3 < \delta.$$

So $y_n$ is in the ball of radius $\delta$ centered at $p$ and so $y_n \in G_{\alpha}$ which is a contradiction.