I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I would like for someone to verify if my proof is technically correct and sufficiently rigorous.
Theorem 6.3.2 Let $(f_n)$ be a sequence of differentiable functions defined on the closed interval $[a,b]$ and assume that $(f_n')$ converges uniformly on $[a,b]$. If there exists a point $x_0 \in [a,b]$ where $f_n(x_0)$ is convergent, then $(f_n)$ converges uniformly on $[a,b]$.
[Abbott 6.3.7] Use the Mean Value Theorem to supply a proof for theorem 6.3.2. To get started observe that the triangle inequality implies that, for any $x \in [a,b]$ and $m,n \in\mathbf{N}$,
\begin{align*} |f_n(x) - f_m(x)| \leq |(f_n(x) - f_m(x))-(f_n(x_0) - f_m(x_0))| + |f_n(x_0) - f_m(x_0)| \end{align*}
Proof.
Pick an arbitrary $\epsilon > 0$.
Fix $x \in [a,b]$ and consider the closed interval $[x,x_0]$ ($[x_0,x]$ if $x_0 < x$).
Since $f_n - f_m$ is continuous on $[x_0,x]$ and differentiable on $(x_0,x)$, by the Mean Value Theorem(MVT), there exists $\alpha \in (x,x_0)$, such that:
\begin{align*} |(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))| &=|f_n'(\alpha) - f_m'(\alpha)|\cdot|x - x_0|\\ &\leq |f_n'(\alpha) - f_m'(\alpha)|\cdot|b-a| \end{align*}
Since $(f_n')$ converges uniformly on $[a,b]$, there exists $N_1 \in \mathbf{N}$ such that for all $n > m \geq N_1$ and for all $x \in [a,b]$,
\begin{align*} |f_n'(x) - f_m'(x)| < \frac{\epsilon}{2 |b-a|} \end{align*}
Since $(f_n(x_0))$ is convergent, by the Cauchy criterion for sequences of real numbers, there exists $N_2(x_0,\epsilon)$, such that for all $n > m \geq N_2$,
\begin{align*} |f_n(x_0) - f_m(x_0)| < \frac{\epsilon}{2} \end{align*}
If we pick $N = \max\{N_1,N_2\}$, then for all $n > m \geq N$ and for all $x \in [a,b]$,
\begin{align*} |f_n(x) - f_m(x)| < \epsilon \end{align*}
Note that, neither $N_1$ nor $N_2$ are functions of $x$, so our choice of $N$ works regardless of what $x$ is. This closes the proof.