Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ and $\ell^s$ for $1 \leq r,s \lt \infty$, unless $r$ and $s$ are both two was discussed (among other things).

There is the somewhat surprising fact that the Banach spaces $X = L^\infty[0,1]$ and $Y = \ell^\infty$ are isomorphic. More precisely, there are mutually inverse bounded linear maps $T: X \to Y$ and $S: Y \to X$ (see below for a proof of existence).

Is there a direct and explicit way to prove this? In other words: I'm wondering whether there is an explicit and natural expression for either $S$ or $T$.

Here's the argument I know: Using Pełczyński's decomposition technique one can prove that $X = L^\infty$ and $Y = \ell^\infty$ are isomorphic as Banach spaces:

Choose a countable partition $[0,1] = \bigcup_{n=0}^\infty E_n$ into disjoint sets of positive measure and send $(x_n)_{n \in \mathbb{N}} \in \ell^\infty$ to $\sum_{n=0}^\infty x_n [E_n]$ to get an isometric embedding $i: Y \hookrightarrow X$. Since $\ell^\infty$ is injective, its image is complemented, in particular, this yields a decomposition $X \cong Y \oplus \widetilde{Y}$.

Choose a dense sequence $(f_n)_{n \in \mathbb{N}}$ of the unit sphere of $L^1[0,1]$. For $h \in L^\infty[0,1]$ let $j(h) = \left( \int f_n \, h \right)_{n \in \mathbb{N}} \in \ell^\infty$ to get an isometric map $j: L^\infty[0,1] \to \ell^\infty$. Since $L^\infty[0,1]$ is injective, its image is complemented in $\ell^\infty$, so this yields a decomposition $Y \cong X \oplus \widetilde{X}$.

Observe that $X \cong X \oplus X$ since $L^\infty[0,1] = L^\infty[0,1/2] \oplus L^\infty[1/2,1] \cong L^\infty [0,1] \oplus L^\infty [0,1]$ and $Y \cong Y \oplus Y$ by decomposing $\mathbb{N}$ into the sets of even and odd numbers. Thus, Pełczyński's argument yields: $$X \cong Y \oplus \widetilde{Y} \cong (Y \oplus Y) \oplus \widetilde{Y} \cong Y \oplus (Y \oplus \widetilde{Y}) \cong Y \oplus X$$ and $$Y \cong X \oplus \widetilde{X} \cong (X \oplus X) \oplus \widetilde{X} \cong X \oplus (X \oplus \widetilde{X}) \cong X \oplus Y$$ so that $X \cong Y \oplus X \cong X \oplus Y \cong Y$.

Of course, one can trace through this argument and “construct” an isomorphism, but the resulting maps are rather messier than what I'm looking for. A further deficit of this argument is that the appeal to injectivity properties makes this inherently non-constructive.

Any simplifications of this argument or pointers to the literature would be welcome.