The second half of the proof is not correct, as the $\delta>0$, does not necessarily work for all $x$, $|x-a|<\delta$, but only for the terms of the sequence you chose.
Updated version. First define $g(a)=\lim_{n\to\infty} g(x_n)$, as you have done. Let $\varepsilon>0$, then there is a $\delta>0$, due to uniform continuity, such that
$$
|x-y|<\delta\quad\Longrightarrow\quad |g(x)-g(y)|<\varepsilon/2.\tag{1}
$$
In particular, let $$|x-a|<\delta/2, \tag{2}$$ and $n_0$, such that $n\ge n_0$, implies $|x_n-a|<\delta/2$. Then every $x$ satisfying $(2)$, also satisfies
$$
|x-x_n|\le |x-a|+|x_n-a|<\delta,\quad \text{for every}\,\,n\ge n_0.
$$
Hence, due to $(1)$, if $x$ satisfies $(2)$ and $n\ge n_0$, then
$$
|g(x)-g(x_n)|<\varepsilon/2,
$$
or equivalently
$$
g(x)-\varepsilon/2<g(x_n)<g(x)+\varepsilon/2, \quad \text{for every $\,\,n\ge n_0\,\,$
and $\,\,|x-a|<\delta/2$,}
$$
and taking the limits, as $n\to \infty$, we obtain
$$
g(x)-\varepsilon/2\le g(a) \le g(x)+\varepsilon/2, \quad \text{for every
$\,\,|x-a|<\delta/2$,}
$$
or
$$
|g(x)-g(a)|\le\varepsilon/2<\varepsilon \quad \text{for every
$\,\,|x-a|<\delta/2$.}
$$
Thus, for every $\varepsilon>0$, there exists a $\delta>0$ such that
$$
|x-a|<\delta/2\quad\Longrightarrow\quad|g(x)-g(a)|<\varepsilon.
$$
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
$I=\mathbb{R}$ unbounded: $f(x)=x$ is uniformly continuous but unbounded.
$I=(0,1)$ bounded, $f(x)=1/x$ is continuous but not uniformly continuous.