$L^1$ and $L^\infty$

Let X be a subspace of Banach space $L^1 ([0,1])$, and for any element in X belongs to $L^\infty ([0,1])$.

We set linear operator
$A : (X,\|\cdot\|_1) \to (L^\infty ([0,1]), \|\cdot\|_\infty)$ as A(f) = f for $f\in X$. I want to prove if X is a closed subspace, then A is continuous.

I tried to use the Closed Graph Theorem. For $f_n, f \in X$ and $g \in L^\infty$, we assume $f_n \to f$ in $L^1$and$A(f_n)\to g$ in $L^\infty$, it is enough to show f = g. However, I cannot move on.