Proving if L, M, N are subspaces of V, then $L \cap M + L \cap N \subseteq L \cap (M+N)$

I understand the intuition behind this statement, however I am having trouble formulating a proof.

Thus far, when attempting to formulate a proof, I can only seem to show that they are equal (below) but this is clearly not always the case.

Any element $k$ from $L \cap M + L \cap N$ will be an element of $L$ as well as an element of $M+N$, thus $k$ is always in both $L$ and $M+N$, i.e. $L \cap (M+N)$ - showing the contrary is the part I am currently struggling with.

As you proved, let $x$ a vector in a first subspace then $x=x_M+x_N$ where $x_M\in L\cap M$ and $x_N\in L\cap N$ so $x_M+x_N\in L$ since $L$ is closed by addition and $x_M+x_N\in M+N$.
Take a vector $x \in L \cap M + L \cap N$.
It means that $x=m+n$ where $m \in L \cap M$ and $n \in L \cap N$. Therefore $m,n \in L$ and as $L$ is a subspace, $m+n \in L$. As clearly $m+n \in M+N$, we get the conclusion.
$M, N\subseteq M+N$, so $L\cap M\enspace(\text{resp. }L\cap N)\subseteq L\cap(M+N)$, whence $$L\cap M+L\cap N\subseteq L\cap(M+N).$$