Show that $\sum^\infty_{n=1} \mu(\{x : |f_n(x) - f(x)| > \epsilon\}) < \infty$ implies $f_n \to f$ a.e.

Your proof is essentially fine. The countability issue is on $\epsilon$. Your proof works for a fixed $\epsilon>0$, but how extend it to all $\epsilon>0$?

Here is how to solve it.

Let $A_{n,m} = \{x : |f_n(x) - f(x)| > \frac{1}{m}\}$ and let $[f_n \nrightarrow f]$ be the set of $x$ where $f_n(x)$ does not converge to $f$. It is easy to see that $$[f_n \nrightarrow f] = \bigcup_{m=1}^\infty\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_{k,m}$$

Since, for all $m$, $\sum^\infty_{n=1} \mu(A_{n,m}) < \infty$, by Borel-Cantelli, we have that $$\mu \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_{k,m} \right)=0$$

So we have $$\mu([f_n \nrightarrow f]) = \mu \left (\bigcup_{m=1}^\infty\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_{k,m}\right)=0$$

So $f_n \to f$ a.e.

Let $A_n = \{x : |f_n(x) - f(x)| \geqslant\epsilon\}$. By Borel Cantelli lemma, we have $$ \sum^\infty_{n=1} \mu(\{x : |f_n(x) - f(x)| \geqslant \epsilon\}) < \infty\implies\mu(\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k)=0 $$ Since $$ x\in\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k\implies x\in A_n \hspace{3 mm}\text{infinitely often} $$ which means that there are infinite $n$ such that $|f_n(x) - f(x)| \geqslant\epsilon$, we have $f_n \to f$ a.e.