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.
Let $(X,\mathcal{A},\mu)$ be a measure space and $(f_n)_{n \in \mathbb{N}}$ such that $f_n \to f$ in measure, i.e.
$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$
for any $\varepsilon >0$. Setting $\varepsilon=2^{-k}$, $k \in \mathbb{N}$, we can choose $n_k$ such that
$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$
for all $n \geq n_k$. Without loss of generality, $n_{k+1} \geq n_k$ for all $k \in \mathbb{N}$. Set
$$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\}.$$
As $$\sum_{k \geq 1} \mu(A_k) \leq \sum_{k=1}^{\infty} 2^{-k} < \infty,$$ the Borel-Cantelli lemma yields
$$\mu \left( \limsup_{k \to \infty} A_k \right) =0.$$
It is not difficult to see that this implies
$$\lim_{k \to \infty} f_{n_k}(x) =f(x)$$
$\mu$-almost everywhere.
Best Answer
Use Cantor's diagonalization argument.