No bounded surjective linear map $L^{\infty} \to L^1$

Let $L^p([0,1])$ denote the $L^p$ space with respect to Lebesgue measure on the unit interval. I want to show there is no bounded surjective linear operator $T \colon L^{\infty}([0,1]) \to L^1([0,1])$.

The open mapping theorem says $T$ is maps open sets to open sets, but I'm not sure how that helps. I know how to prove this if $L^{\infty}$ is replaced by $L^p$ for $1 < p < \infty$ by looking at the adjoint map, but that doesn't work here since $(L^{\infty})^{\ast}$ is not separable

Lemma 1. Let $S:X\to \ell_1$ be surjective bounded operator. Then $X$ contains a complemented subspace isomorphic to $\ell_1$.

proof. $S$ induces an isomorphism $\hat{S}: X/ \text{ker }S\to \ell_1$. So there are $\hat{x}_n\in X$ s.t. $S\hat{x}_n=e_n$ and $||\hat{x}_n+\text{ker } S||\le ||\hat{S}^{-1}||$. It follows that there are $y_n\in \text{ker }S$ s.t. $||\hat{x}_n+y_n||<||\hat{S}^{-1}||+1$. Put $x_n= \hat{x}_n+y_n$. Then $(x_n)$ is equivalent to the canonical base $(e_n)\subset \ell_1$ as follows: for each finitely supported $\xi$, $$||\sum_{n=1}^\infty \xi(n)e_n||\le ||S|| \,||\sum_{n=1}^\infty \xi(n) x_n||\le ||S||(||\hat{S}^{-1}+1)||\sum_{n=1}^\infty \xi(n)e_n||$$The subspace $F=\overline{\text{span}\{x_1,x_2,\dots\}}$ is complemented by composing $S$ with the isomorphism $\varphi:\ell_1\to F, e_n\mapsto x_n$.

Lemma 2. $L_1[0,1]$ contains a complemented subspace isomorphic to $\ell_1$.

proof. Consider a sequence of disjointly supported, normalized sequence $(f_n)\subset L_1[0,1]$.

Lemma 3. If $E$ is an infinite dimensional complemented subspace of $L_\infty$, then $E$ is isomorphic to $L_\infty$.

proof. see Kalton and Albiac's Topics in Banach Space Theory (2nd Edition), 4.3.10 and 5.7.5

Back to the question, suppose that there were some surjective $T:L_\infty[0,1]\to L_1[0,1]$. Let $P:L_1[0,1]\to \ell_1$ be a surjective operator by lemma 2. Then $P\circ T:L_\infty[0,1]\to\ell_1$ is surjective. By Lemma 1, $L_\infty$ contains a complemented copy of $\ell_1$. By Lemma 3, $\ell_1$ would be isomorphic to $L_\infty[0,1]$, contradiction.