We give an approach using the extreme value theorem. The following definition is a key ingredient in the proof: $\newcommand{\diam}{\operatorname{diam}}$
For each $x \in X$, define $h(x) \in \mathbb R^{\geqslant 0}$ to be the infimum of $\diam S$ over all $S \subseteq X$ satisfying the following conditions:
- $x \in S$, and
- $S \nsubseteq U$ for any $U \in \mathcal A$.
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
- $y \in T$;
- $T$ is not contained in any $U \in \mathcal A$;
- $\diam T \leqslant h(y) + \varepsilon$.
Now, consider $S = T \cup \{ x \}$. Clearly,
- $x \in S$;
- $S$ is not contained in any $U \in \mathcal A$ (why?);
- $\diam S \stackrel{\color{Red}{(!!)}}{\leqslant} \diam T + d(x,y) \leqslant h(y) + \varepsilon + d(x,y)$. [Exercise: Justify the inequality marked $\color{Red}{(!!)}$.]
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.
Give $\Bbb N$ the discrete metric:
$$d(m,n)=\begin{cases}1,&\text{if }m\ne n\\0,&\text{if }m=n\end{cases}$$
Clearly this space is not compact, but any positive $d\le 1$ is a Lebesgue number for every open cover of it.
Added: Having given the matter a bit more thought, I can prove the following theorem. Say that a metric space $\langle X,d\rangle$ is Lebesgue if every open cover of it has a Lebesgue number.
Theorem: Let $\langle X,d\rangle$ be a metric space. If $X$ has a non-convergent Cauchy sequence or an infinite closed discrete set of non-isolated points, then $X$ is not Lebesgue. In particular, every Lebesgue space is complete, and every perfect Lebesgue space is compact.
Proof: Suppose first that $\sigma=\langle x_k:k\in\omega\rangle$ is a non-convergent Cauchy sequence in $X$. Let $\langle X^*,d^*\rangle$ be the usual metric completion of $\langle X,d\rangle$, and let $p\in X^*$ be the limit of $\sigma$ in $X^*$. Let $V_0=X^*\setminus B_{d^*}(p,2^{-1})$, and for $k>0$ let $V_k=B_{d^*}(p,2^{-k+1})\setminus \operatorname{cl}_{X^*}B_{d^*}(p,2^{-k-1})$. For $k\in\omega$ let $W_k=X\cap V_k$. Then $\mathscr{W}=\{W_k:k\in\omega\}$ is an open cover of $X$ with no Lebesgue number.
Now suppose that $\{x_k:k\in\omega\}$ is a closed discrete set of non-isolated points in $X$. There is a pairwise disjoint, closure-preserving collection $\{V_k:k\in\omega\}$ such that $x_k\in V_k$ for each $k\in\omega$, so there is a sequence $\langle r_k:k\in\omega\rangle$ of positive real numbers such that $B_d(x_k,r_k)\subseteq V_k$ for each $k\in\omega$, and $\langle r_k:k\in\omega\rangle\to 0$. Let $$W=X\setminus\bigcup_{k\in\omega}\operatorname{cl}_X B_d\left(x_k,\frac{r_k}2\right)\;,$$ and let $\mathscr{W}=\{W\}\cup\{B_d(x_k,r_k):k\in\omega\}$; then $\mathscr{W}$ is an open cover of $X$ with no Lebesgue number. $\dashv$
Best Answer
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.