A dense subset for each of two Banach sets respectively Let $A$ and $B$ be Banach spaces with their own (possibly different) norms. Also, there is a non-empty subset $S \subset A \cap B$ such that $S$ is dense in $A$ and $B$ respectively.
Then, for $x \in A\cap B$, can we always extract a sequence $\{s_n\} \subset S$ such that $s_n \to x$ in $A$ and $s_n \to x$ in $B$?
This question is generalized from the situation $A = L^1(\mathbb{R}^n)$, $B = L^2(\mathbb{R}^n)$ and $S = \mathcal{S}(\mathbb{R}^n)$, in which case, we can find a sequence satisfying the above conditions.
I'd appreciate it if you'd help me!
 A: Here is my proposed counterexample.  It is inspired by considering the dual of the example given in Theorem 2 of http://faculty.missouri.edu/~stephen/preprints/interpolate.html.
Let
$$ Z = L^1([0,1]) \oplus L^1([0,1]) \oplus L^1([0,1]). $$
Let $A$, $B$ be subspaces of $Z$ such that the following norms are finite:
$$ {\|(f,g,h)\|}_{A} = {\|f-g\|}_\infty + {\|g\|}_1 + {\|h\|}_\infty ,$$
$$ {\|(f,g,h)\|}_{B} = {\|f-h\|}_\infty + {\|g\|}_\infty + {\|h\|}_1 .$$
Both spaces are isomorphic to $L^\infty([0,1]) \oplus L^\infty([0,1]) \oplus L^1([0,1])$, so they are Banach spaces.
We can calculate that
$$ {\|(f,g,h)\|}_{A \cap B} := \max\{{\|(f,g,h)\|}_{A},{\|(f,g,h)\|}_{B}\}\approx {\|f\|}_\infty + {\|g\|}_\infty + {\|h\|}_\infty ,$$
because
$$ {\|(f,g,h)\|}_{A \cap B} \le {\|(f,g,h)\|}_{A} + {\|(f,g,h)\|}_{B}
\le 3 ({\|f\|}_\infty + {\|g\|}_\infty + {\|h\|}_\infty) ,$$
and
\begin{align} {\|(f,g,h)\|}_{A \cap B} &\ge \tfrac12({\|(f,g,h)\|}_{A} + {\|(f,g,h)\|}_{B})
\\&\ge \tfrac14{\|f-g\|}_\infty + \tfrac14{\|f-h\|}_\infty + \tfrac12{\|g\|}_\infty + \tfrac12{\|h\|}_\infty
\\&\ge \tfrac14({\|f\|}_\infty-{\|g\|}_\infty) + \tfrac14({\|f\|}_\infty-{\|h\|}_\infty) + \tfrac12{\|g\|}_\infty + \tfrac12{\|h\|}_\infty
\\&\ge \tfrac14 ({\|f\|}_\infty + {\|g\|}_\infty + {\|h\|}_\infty) .\end{align}
Hence
$$ A \cap B = L^\infty([0,1]) \oplus L^\infty([0,1]) \oplus L^\infty([0,1]) .$$
Let
$$ S = C([0,1]) \oplus L^\infty([0,1]) \oplus L^\infty([0,1]). $$
Clearly $S$ is not dense in $A \cap B$.  We show $S$ is dense in $A$, as the argument for $S$ dense in $B$ is essentially identical.
Suppose $x = (f,g,h) \in A$ with ${\|x\|}_A \le 1$, that is,
$$ {\|(f,g,h)\|}_A = {\|f - g\|}_\infty + {\|g\|}_1 + {\|h\|}_\infty \le 1.$$
Note that $f-g\in L^\infty \subset L^1$, and $g\in L^1$, which implies $f \in L^1$.  Let $f_n \in C([0,1])$ be such that ${\|f-f_n\|}_1 \to 0$.
Set
$$ s_n = (f_n, g - f + f_n,h) .$$
Note $g - f + f_n = (g-f) + f_n \in L^\infty([0,1])$, so $s_n \in S$.
Then as $n \to \infty$,
$$ {\|x - s_n\|}_A = {\|(f-f_n, f-f_n, 0)\|}_A = {\|f-f_n\|}_1 \to 0.  $$
