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.
Fix a point $x_0\in (a,b).$ Assume $x\in(x_0,b).$ By using the Lagrange's theorem there exists $c\in(x_0,x)$ such that $f(x)-f(x_0)=f'(c)(x-x_0).$ Thus
$$|f(x)|=|f(x_0)+f'(c)(x-x_0)|\leq |f(x_0)|+|f'(c)||(x-x_0)|\leq |f(x_0)|+M(b-a).$$ Proceeding in the same way you get the bound for $x\in(a,x_0).$ Thus we have shown that the function is bounded.
Edit:
If $x=x_0$ then we have the bound: $|f(x_0)|\le |f(x_0)|+M(b-a).$
Best Answer
Your "proof" could be used for the converse: When $|f'(x)|\leq M$ for all $x\in[a,b]$ then $f$ is Lipschitz continuous on $[a,b]$, and $M$ can serve as a Lipschitz constant.
In fact we are asked to prove that there is an $M>0$ with $|f'(x)|\leq M$ for all $x\in[a,b]$, given that $f$ is Lipschitz continuous on $[a,b]$.
Now when $|f(y)-f(x)|\leq M\>|y-x|$ for all $x$, $y\in[a,b]$ then $$\left|{f(y)-f(x)\over y-x}\right|\leq M\qquad(y\ne x)\ .$$ It follows that for each fixed $x\in[a,b]$ we have $$\left|\lim_{y\to x}{f(y)-f(x)\over y-x}\right|\leq M$$ and therefore $|f'(x)|\leq M$.