One form of Urysohn's lemma (I suspect there may be more than one) is that on a normal space $(A,\mathcal{T})$ with disjoint closed sets $X$ and $Y$, there exists a continuous real valued function $f$ such that $f|_X=0$ and $f|_Y=1$, and $0\leq f\leq 1$ elsewhere on $A$.
I wanted to know if this could be strengthened to finding a uniformly continuous $f$, but results on Google suggested otherwise.
A hint I found online suggested I find a nonconvergent Cauchy sequence $Z=\{z_n\}$ where $z_m\neq z_n$ when $m\neq n$, and take closed sets $X,Y$ with $X\cap Y=\varnothing$ such that $d(x_n,y_n)\to 0$ for $x_n\in X$, $y_n\in Y$ to be subsequences of that Cauchy sequence.
I asked a previous question looking at $(0,1)$ with the usual metric. User Chris Eagle pointed out that the set $X=\{\frac{1}{2n}\mid n\in\mathbb{N}\}$ and $Y=\{\frac{1}{2n+1}\mid n\in\mathbb{N}\}$ are two such closed sets such that $d(x_n,y_n)\to 0$, that is, $(\lim_{n\to\infty}\vert x_n-y_n\vert=0)$ if we take $x_n=\frac{1}{2n}$ and $y_n=\frac{1}{2n+1}$. I took $z_n=\frac{1}{n}$ to be a Cauchy sequence which doesn't converge in $(0,1)$, and can let $X$ and $Y$ be subsequences of $Z$.
Based on this setup, what fails here so that we can not always expect $f$ to be uniformly continuous?