My solution: Consider $f(x)=\sup_n|f_n(x)|$ it is finite everywhere and measurable and so, by Lusin, there is a closed set $F$ s.t $f$ is continuous on $F$ and $\mu([0,1]-F)\leq \epsilon$ Now $F$ is closed and bounded thus compact. Thus $f$ is also bounded uniformly on $F$. Which concludes the proof.
IS this correct? Are there any other nice proofs of this fact?
This question is actually much simpler. There is no need to come up with continuous functions.
Consider $A_N=\{x: \sup_n |f_n(x)| \leq N\}$. These sets increase to $[0,1]$ as $N$ increases to $\infty$ so $m(A_N) \to 1$. This implies that $m(A_N^{c}) <\epsilon$ if $N$ is sufficiently large. Just take $A_{\epsilon} =A_N$ and $B_{\epsilon} =N$.