Given subspaces $V$, $V_{1}$, $V_2$ of $\mathbb{R}^{n}$such that $V_1 \subset V \subset V_1 + V_2$. Is it true $V = V \cap V_{1} + V \cap V_2$?
Could you please verify my proof or propose other ways to solve it?
My proof: since $V_1 \subset V$, $V_1 \cap V = V_1$.
From $V \subset V_1 + V_2$ we know that for $v \in V$, $v_i \in V_i$ the following is true $v = v_1 + v_2$.
Rewrite it as $v - v_1 = v_2$. Hence, $v_2$ should belong to $V$ (meaning $V_2 \subset V)$ and, therefore, $V_1 + V_2 \subset V$.
Combining it with $V \subset V_1 + V_2$, we get $V = V_1 + V_2 = V \cap V_1 + V \cap V_2$.
You reach the wrong conclusion $V_2 \subset V$ as an intermediate step.
You approach can be fixed though: The inclusion $$ V \supset V\cap V_1 + V\cap V_2 $$ is trivial. For the other inclusion, let $v\in V$ and since $V\subset V_1+V_2$ there are $v_1\in V_1\subset V$ and $v_2\in V_2$ such that $v=v_1+v_2$. As you said, this yields $v_2\in V$ and hence $v=v_1+v_2$ with $v_1\in V_1 = V\cap V_1$ and $v_2\in V\cap V_2$. Since this works for any $v\in V$ we get $$ V \subset V\cap V_1 + V\cap V_2. $$ Putting together both inclusions we get the desired identity.