You essentially proved: Every subsequence of $(f_n)$ has a (sub-)subsequence which converges uniformly to zero (why?).
The first bullet point in the link I provided in a comment asserts that then the sequence $(f_n)$ itself must converge to zero.
Here's why: Suppose $f_{n}$ does not converge uniformly to zero. Then there is an $\varepsilon \gt 0$ and a subsequence $f_{n_j}$ with sup-norm $\|f_{n_j}\|_{\infty} \geq \varepsilon$ for all $j$. This subsequence has again a convergent subsequence by Arzelà-Ascoli. Your argument shows that its limit must be zero, hence the subsequence must have uniform norm $\lt \varepsilon$ eventually, contradiction.
Here's the abstract thing:
Let $x_n$ be a sequence in a metric space $(X,d)$. Suppose that there is $x \in X$ such that every subsequence $(x_{n_j})$ has a subsubsequence $(x_{n_{j_k}})$ converging to $x$. Then the sequence itself converges to $x$.
Edit: As leo pointed out in a comment below the converse is also true: a convergent sequence obviously has the property that every subsequence has a convergent subsubsequence.
The proof is trivial but unavoidably uses ugly notation: Suppose $x_n$ does not converge to $x$. Then there is $\varepsilon \gt 0$ and a subsequence $(x_{n_j})$ such that $d(x,x_{n_j}) \geq \varepsilon$ for all $j$. By assumption there is a subsubsequence $x_{n_{j_k}}$ converging to $x$. But this means that $d(x,x_{n_{j_k}}) \lt \varepsilon$ for $k$ large enough. Impossible!
The way this is usually applied in "concrete situations" is to show
- If a subsequence converges then it must converge to a specific $x$. This involves an analysis of the specific situation—this is usually the harder part and that's what you did.
- Appeal to compactness to find a convergent subsubsequence of every subsequence—that's the trivial part I contributed.
I can conclude a subsequence of $f_n$ converges uniformly. But the existence of a uniformly convergent subsequence does not guarantee the original sequence converges uniformly.
Generally, that indeed does not imply the uniform convergence of the full sequence.
However, here we are in a special situation, since we know that the full sequence converges pointwise, and an equicontinuous sequence that converges pointwise converges uniformly on compact sets.
Since the sequence in question is not only equicontinuous but equilipschitz, the proof is easier:
Let $\varepsilon > 0$ be given. Since $\lvert f_n'(x)\rvert \leqslant M$ for all $n$ and $x\in [a,b]$, we have
$$\lvert f_n(x) - f_n(y)\rvert \leqslant \frac{\varepsilon}{4}$$
for all $x,y\in [a,b]$ with $\lvert x-y\rvert \leqslant \frac{\varepsilon}{4M}$.
Choose $N$ large enough that $\frac{b-a}{N} < \frac{\varepsilon}{4M}$, and let $x_k = a + k\frac{b-a}{N}$ for $0 \leqslant k \leqslant N$. For each $k$, there is an $n_k \in \mathbb{N}$ such that $\lvert f_n(x_k) - f(x_k)\rvert < \frac{\varepsilon}{4}$ for all $n \geqslant n_k$. Let $n(\varepsilon) = \max \{ n_k : 0 \leqslant k \leqslant N\}$.
Then, for every $x \in [a,b]$ and $n, m \geqslant n(\varepsilon)$ we have
$$\begin{align}
\lvert f_n(x) - f_m(x)\rvert &\leqslant \lvert f_n(x) - f_n(x_k)\rvert + \lvert f_n(x_k) - f(x_k)\rvert + \lvert f(x_k) - f_m(x_k)\rvert + \lvert f_m(x_k) - f_m(x)\rvert\\
&\leqslant \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \varepsilon,
\end{align}$$
for $k = \left\lfloor N\frac{x-a}{b-a}\right\rfloor$, so $\lvert x-x_k\rvert \leqslant \frac{\varepsilon}{4M}$.
Thus, for every $\varepsilon > 0$, we found an $n(\varepsilon)\in\mathbb{N}$ with
$$\sup \{ \lvert f_n(x) - f_m(x)\rvert : x \in [a,b], \, n,m \geqslant n(\varepsilon)\} \leqslant \varepsilon,$$
i.e. the sequence converges uniformly.
Best Answer
Let $f$ be the common limit of the uniformly convergent subsequence. Suppose $f_n$ does not converge uniformly to $f$, so we have some $\epsilon>0$ and a collection of points $x_{n_k}\in [a,b]$ such that $n_k\to \infty$ and $\|f_{n_k}(x_{n_k})-f(x_{n_k})\|>\epsilon$. But then no subsequence of $f_{n_k}$ can converge uniformly to $f$, a contradiction. Hence $f_n$ converges uniformly to $f$.