The intuition is the following : for each $x$ of $X$ is included in at least one $U$ of $\mathcal U$. Since $U$ is open, it contains a ball centered at $x$ and with positive radius, right ?
Let $r(x, U)$ denote the supremum of all such radius. And let $r(x)$ the supremum of all the $r(x,U)$, with $U\in\mathcal U$.
So $r(x)$ is a continuous positive function on $X$. But if $X$ is not compact, you may find a sequence $(x_n)$ such that $r(x_n)$ tends to zero.
However, if your metric space $X$ is compact, then $r$ shall have a minimum. This minimum is of course positive since $r$ is. This minimum is a Lebesgue number, the greater one.
And in fact, the intuition is a proof !
A covering with positive Lebesgue number is called a uniform covering.
We are going to prove it by contradiction. Suppose $X$ doesn't satisfy the Lebesgue Number Lemma, i.e.
There exists an open covering $\{U_{\alpha}\}_{\alpha\in J}$ such that
for all $\delta>0$, there exists $A_{\delta}$, diam $A_{\delta}<\delta$ and $A_{\delta}\subsetneq U_{\alpha}$ for all $\alpha\in J$.
It suffices to prove that there exists $E= \{x_{1},y_{1},x_{2},y_{2},\dots\}$ such that $E$ has no limit point and $d(x_n,y_n) \to 0$ when $n\to \infty$.
Because $E$ has no limit points, it follows that $E $ is closed in $X.$ In its subspace topology, $E$ is discrete. Hence any function defined on $E$ is continuous. Let's take $g = 0$ on $\{x_{n}\},$ $g = 1$ on $\{y_{n}\}$. By the Tietze extension theorem (since $E$ is closed), $g$ extends to a function $G:X\to \mathbb {R}$ that is continuous on $X$. But $G$ is not uniformly continuous (As $d(x_{n},y_{n}) \to 0,|G(x_{n})-G(y_{n})|=1$). Contradictory to the starting assumption that every continuous function on $X$ is uniformly continuous.
How to obtain such $E$? Simply choose $x_n,y_n\in A_{1/n}$ with $x_n\neq y_n$. I claim that no subsequence $x_{n_k}$ can converge in $X$. If $x_{n_k}\to x$, then $x\in U_\alpha$ for some $\alpha$. Because $U_\alpha$ is open, there exists $\epsilon>0$ such that the open ball $B(x,\epsilon)$ of radius $\epsilon$ centered at $x$ is contained in $U_\alpha$. Since $x_{n_k}\to x$, every neighboorhood of $x$ contains infinitely many $x_{n_k}$. Choose $n_k$ large enough so that $1/{n_k}<\epsilon$. Then $A_{n_k}\subset B(x,\epsilon) \subset U_{\alpha}$. Contradiction. Similarly, no subsequence $y_{n_k}$ can converge in $X.$
Now if $E$ has a limit point, a subsequence $x_{n_k}$ or $y_{n_k}$ must converge in $X.$ Hence it's our desired $E$.
Best Answer
We give an approach using the extreme value theorem. The following definition is a key ingredient in the proof: $\newcommand{\diam}{\operatorname{diam}}$
We now study the map $h: X \to \mathbb R^{\geqslant 0}$ as defined above. Note first that $h(x) > 0$ for every $x \in X$. [Proof is left as exercise.]
Lipschitzness. The main technical idea is to show that $h$ is $1$-Lipschitz. Fix any $x, y \in X$; we want to show that $h(y) \geqslant h(x) - d(x,y)$. Further fix an arbitrary $\varepsilon > 0$. By the definition of $h(y)$, there exists $T \subseteq X$ such that
Now, consider $S = T \cup \{ x \}$. Clearly,
Therefore, by definition of $h(x)$, we can see that $h(x) \leqslant h(y) + d(x,y) + \varepsilon$. Since this is true for all $\varepsilon > 0$, it follows that $h(x) \leqslant h(y) + d(x,y)$.
Wrap-up of the proof. Since $h$ is Lipschitz, it is also continuous on $X$. Furthermore, being a continuous function over a compact set, $h$ is guaranteed (by the extreme value theorem) to attains its minimum over $X$, and this minimum is strictly positive. Let $\delta > 0$ be any number that is strictly smaller than $h(x)$ for all $x \in X$. It only remains to check that such a $\delta$ satisfies the requirements of the problem. I leave that as a simple exercise.