We assume there is no $\varepsilon>0$. For every $n\in\mathbb{N}$ we define $\varepsilon_n:=\frac{1}{n}$. Now it exists for every $\varepsilon_n$ a $x_n\in X$ such that $K_{\varepsilon_n}(x_n)\nsubseteq U$ for all $U\in\mathfrak{U}$ by assumtion. The sequence $(x_n)_{n\in\mathbb{N}}$ has a convergent subsequence, such that $x_{n_\nu}\to x$ for $\nu\to\infty$. Now, we choose $U\in\mathfrak{U}$ such that $x\in U$. $U$ is open, hence there exists a $\delta>0$ such that $K_\delta(x)\subseteq U$. Because our subsequence converge there is a $N\in\mathbb{N}$ such that $x_{n_\nu}\in K_{\frac{\delta}{2}}(x)$ for all $\nu\geq N$. Let $\varepsilon:=\frac{\delta}{2}$. So we get $K_\varepsilon(x_{n_\nu})\subseteq K_\delta(x)\subseteq U$ für all $\nu\geq N$ However, $\varepsilon_n$ is convergent with $\lim_{n\to\infty}\varepsilon_n=0$ and $K_\varepsilon(x)$ is open. Hence there exists a $N'\geq N$ such that $K_{\varepsilon_n}(x_n)\subseteq K_\varepsilon(x)\subseteq U$, this is our contradiction.
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
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.
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. $\dashv$