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!
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 $n \to \infty$, $$ {\|x - s_n\|}_A = {\|(f-f_n, f-f_n, 0)\|}_A = {\|f-f_n\|}_1 \to 0. $$