Let $E$ be as supposed. The natural inclusion $T : L^\infty([0,1]) \hookrightarrow L^2([0,1])$ is bounded, so $E = T^{-1}(E)$ is therefore also closed in $L^\infty$. By the open mapping theorem, it follows that $T^{-1}$ is bounded on $E$, so there exists $C$ such that for all $f \in E$, $||f||_\infty \le C ||f||_2$. Now if $||f||_2 = 1$, we have $||f||_\infty \le C$, and so by noting $$1 = \int |f|^2 \le C^2 m(|f| > \epsilon) + \epsilon^2$$ and taking, say, $\epsilon = 1/2$, we have $m(|f| > 1/2) \ge 1/4C^2$.
Now suppose $E$ is infinite dimensional; then it contains an $L^2$-orthonormal sequence $\{f_n\}$. By replacing $f_n$ by $-f_n$ as necessary we may assume that for each $f_n$, $m(f_n > 1/2) \ge 1/8C^2$. By a pigeonhole argument there is a set $A$ of positive measure where $f_{n_k} > 1/2$ for infinitely many $n_k$. n_k$(edit: actually I am not sure about this step). Now$f = \sum_k k^{-1} f_{n_k}$converges in$L^2$and so$f \in E$; since the$L^2$and$L^\infty$norms are equivalent on$E$, the sum also converges uniformly to$f$on a set of full measure. But this implies that$f = +\infty$a.e. on$A$, which is absurd. 1 Here's one solution. There may be cleaner ones. Let$E$be as supposed. The natural inclusion$T : L^\infty([0,1]) \hookrightarrow L^2([0,1])$is bounded, so$E = T^{-1}(E)$is therefore also closed in$L^\infty$. By the open mapping theorem, it follows that$T^{-1}$is bounded on$E$, so there exists$C$such that for all$f \in E$,$||f||_\infty \le C ||f||_2$. Now if$||f||_2 = 1$, we have$||f||_\infty \le C$, and so by noting $$1 = \int |f|^2 \le C^2 m(|f| > \epsilon) + \epsilon^2$$ and taking, say,$\epsilon = 1/2$, we have$m(|f| > 1/2) \ge 1/4C^2$. Now suppose$E$is infinite dimensional; then it contains an$L^2$-orthonormal sequence$\{f_n\}$. By replacing$f_n$by$-f_n$as necessary we may assume that for each$f_n$,$m(f_n > 1/2) \ge 1/8C^2$. By a pigeonhole argument there is a set$A$of positive measure where$f_{n_k} > 1/2$for infinitely many$n_k$. Now$f = \sum_k k^{-1} f_{n_k}$converges in$L^2$and so$f \in E$; since the$L^2$and$L^\infty$norms are equivalent on$E$, the sum also converges uniformly to$f$on a set of full measure. But this implies that$f = +\infty$a.e. on$A\$, which is absurd.