Differentiable function with bounded derivative, yet not uniformly continuous

Let us start with

$$h(x) = \begin{cases} \hphantom{-}4(x+2) &, -2\leqslant x < -\frac{3}{2}\\ -4(x+1) &, -\frac{3}{2} \leqslant x < -1\\ -4(x-1) &, \hphantom{-}\; 1 \leqslant x < \frac{3}{2}\\ \hphantom{-} 4(x-2) &, \hphantom{-}\frac{3}{2}\leqslant x < 2\\ \qquad 0 &, \hphantom{-} \text{ otherwise.} \end{cases}$$

For $c > 0$, let

$$h_c(x) = c\cdot h(c\cdot x).$$

Then $h_c$ is continuous, and $\int_{-\infty}^0 h_c(x)\,dx = 1$ as well as $\int_{-\infty}^\infty h_c(x)\,dx = 0$. Now let

$$g(x) = \sum_{n=1}^\infty h_{5^n}\left(x-n-\frac{1}{2}\right).$$

Every $g_n(x) = h_{5^n}\left(x-n-\frac12\right)$ vanishes identically outside the interval $[n,n+1]$, so $g$ is continuous, and

$$f(x) = \int_0^x g(t)\,dt$$

is well-defined and continuously differentiable.

Furthermore, $f(x) \equiv 0$ on every interval $\left[n, n+\frac{1}{2} - \frac{2}{5^n}\right]$, and $f(x) \equiv 1$ on every interval $\left[n+\frac{1}{2}-\frac{1}{5^n}, n+\frac{1}{2}+\frac{1}{5^n}\right]$. Thus on

$$X = \bigcup_{n=1}^\infty \left(\left[n, n+\frac{1}{2} - \frac{2}{5^n}\right] \cup \left[n+\frac{1}{2}-\frac{1}{5^n}, n+\frac{1}{2}+\frac{1}{5^n}\right]\right)$$

we have $f' \equiv 0$, so the derivative is bounded, but

$$f\left(n+\frac{1}{2}-\frac{1}{5^n}\right) - f\left(n+\frac{1}{2}-\frac{2}{5^n}\right) = 1$$

for all $n$, while the distance between the two points is $5^{-n}$ which becomes arbitrarily small, so $f$ is not uniformly continuous on $X$.

If the sentence

Note that $X$ cannot be an interval, a disjoint union of intervals, nor can it be a discrete set.

was meant to forbid a construction as above where $X$ is a disjoint union of intervals, we can obey the letter of the law (but not the spirit) by adding an arbitrary subset of $(-\infty,0)$ that is not a union of disjoint intervals.