Dense subset of two Banach spaces also dense in the intersection

I have a negative answer for your "easier" question based on the construction of https://mathoverflow.net/a/184471.

Let $X_1 := (X, \|\cdot\|_1)$ be an infinite dimensional Banach space and let $\varphi$ be an unbounded linear functional on $X_1$. We fix $y \in X$ with $\varphi(y) = 1$ and define $$ S x := x - 2 \, \varphi(x) \, y.$$ It is easily checked that $S^2 x := S S x = x$. The norm $$ \|x \|_2 := \| S x\|_1$$ gives rise to the normed space $X_2 := (X, \|\cdot\|_2)$. Since $S :X_2 \to X_1$ is an isometric isomorphism (by definition), $X_2$ is complete.

From $\varphi(x) = -\varphi(Sx)$ one can check that $\varphi$ is also unbounded on $X_2$. Indeed, we find $x_n \in X$ with $\varphi(x_n) \ge n$ and $\|x_n\|_1=1$. Hence, $\varphi( S x_n) \ge n$ and $\|S x_n\|_2 = \|x_n\|_1 = 1$.

Thus, the kernel of $\varphi$ is dense in $X_1$ and $X_2$.

However, we can check that $\varphi$ is bounded w.r.t. $\|\cdot\|=\|\cdot\|_1+\|\cdot\|_2$: $$ 2 \, \|y\|_1 \, |\varphi(x)| = \| 2 \, \varphi(x) \, y \|_1 \le \|x\|_1 + \| x - 2 \, \varphi(x) \, y \|_1 = \|x\|_1 + \| S x\|_1 = \|x\|.$$ Hence, the kernel of $\varphi$ is closed and therefore not dense w.r.t. the norm $\|\cdot\|$ in $X$.

I would imagine that your original question would be also interesting if we add the following (reasonable) assumption: if $\{z_n\} \subset X \cap Y$ satisfies $z_n \to x$ in $X$ and $z_n \to y$ in $Y$ then $x = y$. Note that this is not satisfied in my counterexample.