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 $n \to \infty$,
$$ {\|x - s_n\|}_A = {\|(f-f_n, f-f_n, 0)\|}_A = {\|f-f_n\|}_1 \to 0. $$
If $X=E\times F$, let $p:X\to E$ be the projection $p:(x,y)\longmapsto x$. Let $V\subset E$ be open. Then
$$
p^{-1}(V)=V\times F.
$$
Because the topology is the product topology, $V\times F$ is open. Hence $p$ is continuous.
This means that if your projection $P$ is not continuous, the direct sum $\operatorname{ran}P\oplus\ker P$ is not topological.
It is easy to produce unbounded projections. Indeed, given any infinite-dimensional Banach space $X$, let $g$ be an unbounded linear functional. Fix nonzero $z\in X$ and define
$$
Px=g(x)z.
$$
Then $P$ is an unbounded rank-one projection, and $\ker P=\ker g$ is dense.
Best Answer
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.