Here is a simple proof that is tediously long. As with many proofs dealing
with measure, it proceeds by showing an equivalent result for indicator
functions, then simple functions and then measurable function.
I think the essence of the result is 1), the rest is generalisation.
I am assuming that you are using the Lebesgue measure $m$.
Since we can extend $f$ by letting $f(x)=0$ for $x \notin E$, there is no loss of generality in taking $E= \mathbb{R}$.
1) If $A$ has finite measure and $\epsilon>0$, we can find a finite collection of disjoint open intervals $I_1,...,I_n$ such that $m (A \triangle \cup_{k \le n} I_k) \le \epsilon$.
Suppose $A$ is measurable set of finite measure. Then for any $\epsilon>0$ there is a (possibly countable) collection of open intervals $\{I_k\}$ such that $A \subset \cup_k I_k$ and $mA \le \sum_k m I_k \le mA+ {\epsilon \over 2}$. Since $m (\cup_k I_k) = m A + m (\cup_k I_k \setminus A)$, we have
$m ( \cup_{k \le n} I_k \setminus A) \le m ( \cup_{k} I_k \setminus A) \le { \epsilon \over 2}$, for any $n$.
We also have $mA = m (A \cap \cup_{k \le n} I_k ) + m ( A \setminus \cup_{k \le n} I_k)$, hence by continuity of measure, we can find some $n$ such that
$m ( A \setminus \cup_{k \le n} I_k) \le { \epsilon \over 2}$. Hence for some $N$ we have $m (A \triangle \cup_{k \le n} I_k) \le \epsilon$.
To obtain disjoint intervals, suppose two of the open intervals $I_1,...,I_n$ overlap, say $I_{k_1}$ and $I_{k_2}$. Then $I_{k_1} \cup I_{k_2}$ is also an open interval that contains both intervals and
$m (I_{k_1} \cup I_{k_2}) \le m I_{k_1} + m I_{k_2}$. So, we remove
$I_{k_1}$, $I_{k_2}$ from the collection and replace them by the combined
interval $I_{k_1} \cup I_{k_2}$. We continue until there are no overlapping intervals. Let $\{ I_k' \}_{k=1}^{n'}$ be the resulting collection of disjoint intervals. Since $\cup_{k=1}^n I_k = \cup_{k=1}^{n'} I_k'$, all of the above estimates are still valid and we have
$m (A \triangle \cup_{k \le n'} I_k') \le \epsilon$.
2) If $B = I_1 \cup \cdots \cup I_n$ where the $I_k$ are disjoint intervals, then $1_{B} = \sum_{k=1}^n 1_{I_k}$ is a step function.
3) Let $A$ have finite measure and $\epsilon>0$, then there exists a step function $s$ and a set $\Delta$ such that $1_A(x) = s(x)$ for $x \notin \Delta$ and $m \Delta \le \epsilon$. Furthermore, if $A \subset J$, where $J$ is an bounded interval, we can take $\Delta \subset J$ as well (in particular, we have
$s(x) = 0$ for $x \notin J$).
From 1), we have disjoint open intervals $I_1,...,I_n$ such that
$m (A \triangle \cup_{k \le n} I_k) \le \epsilon$. Let $s=\sum_{k=1}^n 1_{I_k}$ and $\Delta = A \triangle \cup_{k \le n} I_k$.
Then we see that $m \Delta \le \epsilon$ and
if $x \notin \Delta$, we have $s(x) = 1_A(x)$.
If $A \subset J$, a bounded interval, let $s'=s \cdot 1_J$, and note that $s'$
is a step function. Notice that if $x \notin J$, then $s'(x) = 0 = 1_A(x)$ and so we have $s'(x) = 1_A(x)$ for $x \notin \Delta' = \Delta \cap J$, with $m \Delta' \le m \Delta \le \epsilon$.
4) If $A$ is measurable and $\epsilon>0$, then there exists a step function $s$ and a set $\Delta$ such that $1_A(x) = s(x)$ for $x \notin \Delta$ and $m \Delta \le \epsilon$.
(Note that the finite sum of step functions is a step function.)
Let $B_n = A \cap [n,n+1)$ and $\epsilon>0$. Let $s_n$ and $\Delta_n \subset [n,n+1)$ be the step function and set such that $1_{B_n}(x) = s_n(x)$ for all $x \notin \Delta_n$ with $m \Delta_n \le {1 \over 2^{|n|+2}} \epsilon$. (Note that we can take $J=[n,n+1)$, so that $\Delta_n \subset [n,n+1)$. Consequently, $s_n(x) = 0$ when $x \notin [n,n+1)$.)
Let $s = \sum_n s_n$ and $\Delta = \cup_n \Delta_n$. Then if $x \notin \Delta$, we have $s(x) = \sum_n s_n(x) = \sum_n 1_{B_n}(x) = 1_{B_{\lfloor x \rfloor}} (x) = 1_A(x)$ and $m \Delta \le { 3 \over 4} \epsilon$.
5) If $\sigma$ is a real valued simple function and $\epsilon>0$, there exists a step function $s$ and a set $\Delta$ such that $m \Delta \le \epsilon$ and
$\sigma(x) = s(x) $ for $x \notin \Delta$.
Suppose $\sigma = \sum_{k=1}^n \alpha_k 1_{A_k}$, where the $A_k$ are measurable. Using 4), choose $s_k, \Delta_k$ such that $1_{A_k}(x) = s_k(x)$ for $x \notin \Delta_k$, and $m \Delta_k \le {1 \over n} \epsilon$. If we let
$s = \sum_k \alpha_k s_k$ (note that $s$ is a step function)
and $\Delta = \cup_k \Delta_k$, we have $m \Delta \le \epsilon$ and if
$x \notin \Delta$, then $\sigma(x) = s(x)$.
And finally:
6) If $f$ is measurable, there are step functions $s_n$ such that $s_n(x) \to f(x)$ for ae. $x$.
There are real valued simple functions $\sigma_n$ such that $\sigma_n(x) \to f(x) $ for all $x$. Using 5), choose step functions $s_n$ and sets $\Delta_n$
such that $s_n(x) = \sigma_n(x)$ for $x \notin \Delta_n$, and
$m \Delta_n \le {1 \over n} {1 \over 2^{n+1}}$.
Note that if $s_n(x)$ does not converge to $f(x)$, then we must have $s_n(x) \neq \sigma_n(x)$ for infinitely many $n$. This means $x \in \Delta_n$ infinitely often. Equivalently, we have
$x \in Q=\cap_n \cup_{k \ge n} \Delta_k$.
Since $m Q \le m (\cup_{k \ge n} \Delta_k)
\le \sum_{k \ge n} {1 \over k 2^{k+1}} \le {1 \over n}$ for all $n$, we see that $m Q = 0$.
Consequently, for any $x \notin Q$, there is some $N$ such that
$s_n(x) = \sigma_n(x)$ for all $n \ge N$, and so
$s_n(x) \to f(x)$.
Best Answer
The end of your proof is incorrect. The problem is that you have proven that for each $x$ and $\varepsilon$ there is a $k$ with $a < f_k(x) + \varepsilon$. But that does not imply that there is a $k$ with $a < f_k(x)$. For example, consider the case where $a = 0$ and $f_k \equiv 0$ for all $k$. Then $a < f_k(x) + \varepsilon$ for every $\varepsilon > 0$, but there is no $k$ with $f_k(x) > a$.
Instead, you need to choose a particular value for $\varepsilon$. The trick is that because the supremum is strictly larger than $a$, there must be a $k$ such that $f_k(x)$ is squeezed between $a$ and the supremum. By choosing $\varepsilon = \sup f_n(x) - a > 0$ you obtain that there is a $k$ such that $$f_k(x) > \sup f_n(x) - a = \sup f_n(x) - (\sup f_n(x) - a) = a.$$