I want to prove this theorem:
Let $(f_n)$ be a sequence of differentiable functions defined on the closed interval $[a, b]$, and assume $(f_n')$ converges uniformly on[a, b]. If there exists a point $x_0 ∈[a, b]$ where $f_n(x_0)$ is convergent, then $(f_n)$ converges uniformly on $[a, b]$.
The text suggests to use the MVT:
Let $x \in [a, b]$ and assume, without loss of generality, that $x > x_0$ . Applying the Mean Value Theorem to the function $f_n -f_m$ on the interval $[x_0 , x]$, we get that there exists a point $\alpha$ such that $(f_n(x) - f_m(x)) - (f_n (x_0) - f_m(x_0))$ = $(f'_n(\alpha)- f'_m(\alpha))(b - a) \space \space \space \space \space \space \space \space \space \space \space \space(1)$
My question is that, since we are using the MVT over an arbitrarily constructed interval $[x_0,x]$, does this not suggest that $(1)$ should be multiplied by $(x-x_0)$ as opposed to $(b-a)$?
And also, just wanted to confirm (whilst seemingly obvious, however not stated in the text) if it is true that $\alpha \in [x_0,x]$
You are right in both of your suspicions (though precisely speaking, $\alpha\in]x,x_0[$. But that's not too big of a deal, since you want to bound $f_n(x)-f_m(x)$:
$$\begin{align*} |f_n(x)-f_m(x)|&=|(f_n(x_0)-f_m(x_0))+(f_n'(\alpha)-f_m'(\alpha))(x-x_0)| \\&\leq |f_n(x_0)-f_m(x_0)|+|f_n'(\alpha)-f_m'(\alpha)||x-x_0| \\&\leq |f_n(x_0)-f_m(x_0)|+|f_n'(\alpha)-f_m'(\alpha)|M \end{align*}$$
where $M:=|b-a|.$
The first summand in the last expression can be made arbitrarily small since $\{f_n(x_0)\}_n\subseteq\mathbb{R}$ is convergent. The second summand too can be made arbitrarily small by the uniform convergence of the derivative. Hence we are done (I'll leave it to you to determine which epsilon(s) should be chosen for a clean exposition).