Question: Let $f\_n: [0, 1] \to \mathbb{R}$ be a bounded sequence of functions in $L^2([0, 1])$. This means that there exists $C\_0 > 0$ so that$$\|f\_n\|\_{L^2} \le C\_0 \text{ for all }n.$$Assume that $f\_n$ converges to $f$ in $L^1([0, 1])$. I have two questions.

1. Do we have necessarily have that $f \in L^2([0, 1])$?
2. Do we necessarily have that $\|f\|\_{L^2} \le C\_0$?

Answer: 
Here is a proof inspired by @Bananach's answer: (I wanted to find a proof that didn't utilise pointwise limits to show equality.)

The set $\bar{B}(0,C\_0) \subset L^2[0,1]$ is weakly compact (by Banach Alaoglu),
hence there is some $\tilde{f} \in \bar{B}(0,C\_0)$ such that $f\_{k\_k}\overset{\text{weak}}{\to} \tilde{f}$ for some subsequence.

To finish, we just need to show that $f = \tilde{f}$.

Suppose $\phi$ is bounded (that is, there is some $B$ such that $|\phi(x)| \le B$ for all $x$) and measurable, then we have (I am implicitly using
the fact that $L^2[0,1] \subset L^1[0,1]$):
$|\langle \phi , f -\tilde{f} \rangle | \le |\langle \phi , f -f\_{n\_k} \rangle | + |\langle \phi , \tilde{f}-f\_{n\_k} \rangle | \le \|\phi\|\_\infty \|f -f\_{n\_k} \|\_1 + |\langle \phi , \tilde{f}-f\_{n\_k} \rangle |$
and hence
$\langle \phi , f -\tilde{f} \rangle = 0$ for all bounded measurable $\phi$.

By choosing $\phi$ to be a suitable indicator function, we see that $f(x) = \tilde{f}(x)$ ae. $x$.
Answer to both questions is yes. Bounded sequences in $ L^2$ have weakly converging subsequences. By uniqueness of the almost everywhere pointwise limit, these weak limits coincide with your $ f $. Furthermore, the norm of the weak limit of a subsequence is smaller than the limit inferior of the norms of the elements of the sequence.

Using a bit less abstract arguments, define a functional $\phi$ on $ L^2$ as follows. For each $ g\in C [0,1]\cap L^2[0,1] $ define $\phi(g ) $ as $\int f g $. Check that the operator norm of $\phi$, on the subspace where it ia defined equipped with the $ L^2$ norm, is smaller than $ C\_0$. Therefore we may extend $\phi $ to the full space, and conclude by observing that the $ L^2$ norm of $ f $ equals the operator norm of that extension, which is still bounded by the same constant. 

As I said, only a bit less abstract, and maybe with more details to be filled.