If $X=M\oplus N$, is it true $(A\cap M)\oplus (A\cap N)\subseteq A$?

Question

Let $X$ be a norm space and $X=M\oplus N$ where $M,N$ are closed subspace of $X$. Take a closed set $A\subseteq X$.

I know that $(A\cap M)\oplus (A\cap N)\neq A$, for example consider $X=\mathbb{R}^2$, $M=\mathbb{R}\times \{0\}$, and $N=\{0\}\times \mathbb{R}$. Then if $A= \Delta_X=\{(x, x):x\in\mathbb{R}\}$,then $(A\cap M)\oplus (A\cap N)=\{(0,0)\}$

Is it true that $(A\cap M)\oplus (A\cap N)\subseteq A$? Please help me to know it.

Answer 1

Counter-example: $X=\mathbb R^{2}, M=\{(x,0): x \in \mathbb R\},N=\{(0,y): y \in \mathbb R\}, A =M \cup N$. [$(1,0)+(0,1)\notin A$].