Yes, $f(x)=x$ whereas $f'(x)=1$.
Having seen the competition in here answering with the simplest possible function, why not define the function
$$
f(x)=\int\sin(\ln(x))dx=-\frac{1}{2}x\cdot\left(\cos(\ln(x))-\sin(\ln(x))\right)
$$
with its derivative
$$
f'(x)=\sin(\ln(x))\in[-1,1]
$$
Since $x$ is not bounded and $\cos(t)-\sin(t)=1$ whenever $t$ is a multiple of $2\pi$ and the equation $\ln(x)=t$ has solutions for all $t$ we see that $f(x)=-\frac{1}{2}x$ for an unbounded set of $x$-values thus being unbounded.
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.
Best Answer
Note that it is crucial the domain of the function be a bounded open interval, because if the domain is $\Bbb{R}$ then the claim is false ($f(x)=x$ is unbounded on $\Bbb{R}$ yet its derivative is constant hence trivially bounded).
One way of phrasing the argument is contrapositively, and I find that simpler. So, we suppose $f'$ is bounded on $(a,b)$, say by $B$, and we want to show $f$ itself is bounded. So, fix an $x_0\in(a,b)$ (for example $\frac{a+b}{2}$... it doesn't really matter, just pick a point and keep it fixed for the rest of the discussion), and let $x\in (a,b)$ be arbitrary. Then, \begin{align} |f(x)-f(x_0)|\leq B|x-x_0|. \end{align} (if $x=x_0$ this inequality is trivially true, if $x<x_0$ use the mean-value theorem on the interval $(x,x_0)$ while if $x_0<x$ then use the mean-value theorem on $(x_0,x)$). In particular, by the reverse triangle inequality, \begin{align} |f(x)|&\leq |f(x_0)|+B|x-x_0|\\ &\leq |f(x_0)|+B(b-a). \end{align} Since $x\in (a,b)$ is arbitrary, we have that $f$ is bounded by $M:=|f(x_0)|+B(b-a)$.
The crux of this argument is that if $f'$ is bounded, then $f$ satisfies a Lipschitz condition, and therefore is bounded on every bounded interval.