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.
As stated, the answer to your question is no. The Cantor function is a common counterexample when the derivative is required to exist only almost everywhere. It is continuous, has zero derivative a.e., in particular in an open set of full measure, but it is not Lipschitz continuous, nor absolutely continuous. You can make an analogous example in the cube $\mathbb{R}^n$ considering the Cantor function of the first variable.
If instead you take a function $f\in L^1_{loc}(\mathbb{R}^n)$ and require its gradient to exist in the distributional sense and to be represented by a function in $L^\infty$, then you are by definition stating that $f$ belongs to the Sobolev space $W^{1,\infty}(\mathbb{R}^n)$, which can be proven to coincide with the space of Lipschitz functions, see for example this question. This still holds if you replace $\mathbb{R}^n$ with a convex set $\Omega$ (there are some weaker conditions on $\Omega$ under which it works, but it doesn't work with arbitrary domains). Note that you don't need to assume a priori that $f$ is continuous, nor that the gradient is continuous almost everywhere.
Best Answer
The last part of your argument is not clear for me, You can remark that since $f$ is uniformly continuous on $(a,b)$, you can extend it by continuity to $[a,b]$ and a continuous function on a closed interval is bounded,
another approach
Suppose that there exists $M>0$ such that $|f'(x)|<M$, for every integer $n$, there exists $x_n,y_n$ such that $|f(x_n)-f(y_n)|>n$, we have $|f(x_n)-f(y_n)|=|f'(c_n)||x_n-y_n|\leq M|b-a|$, this implies that every integer $n$ verifies $n\leq M|b-a|$. Contradicition.