show $(V_1 \cap V_3)+(V_2 \cap V_3)=(V_1+V_2)\cap V_3$ if $ V_1 \subset V_3$ Let $V_{1,2,3}$ be Vectorspaces.
We want to show the equation above.
My attempt:
\begin{align*}
(V_1 \cap V_3)+(V_2 \cap V_3)
& =\left \{ v: \exists u_1 \in V_1 , \ u_2 \in V_2 \cap V_3, \ v = u_1 + u_2 \right \} \\
& = \left \{v : \exists u_1 \in V_1, \ (u_2 \in V_2 \wedge u_2 \in V_3), \ v = u_1 + u_2  \right \} \\
& = \left \{ v : \exists u_1 \in V_1, \ u_2 \in V_2, \ v = u_1 + u_2 \right \} \\
& \qquad \qquad \cap
\left \{ v: \exists u_1 \in V_1, \ u_2 \in V_3, \ v = u_1 + u_2 \right\} \\
& = (V_1 + V_2)\cap(V_1 + V_3) \\
& \overset{V_1 \subset V_3}{=} (V_1+V_2) \cap V_3
\end{align*}
I have a serious doubt that I'm missing something here. Any help is much appreciated.

I think your main problem is getting bogged down in symbols, a few words here and there can really make things easier to write and understand.

Let's show both inclusions separately.
First we will show that $(V_1 + V_2)\cap V_3 \subseteq (V_1 \cap V_3) + (V_2 \cap V_3)$.
Take $v \in (V_1 + V_2)\cap V_3$. Then it can be written as $v = v_1 + v_2$ for some $v_1\in V_1$ and $v_2 \in V_2$.
Actually we find that $v_1,v_2\in V_3$ automatically. Why? Well $v_1\in V_1 \subset V_3$ and $v\in V_3$, hence $v_2 = v - v_1 \in V_3$ since $V_3$ is a vector space.
Thus $v \in (V_1 \cap V_3) + (V_2 \cap V_3)$.

Now for the other inclusion, $(V_1 \cap V_3) + (V_2 \cap V_3)\subseteq (V_1 + V_2)\cap V_3 $. 
Take $v \in (V_1 \cap V_3) + (V_2 \cap V_3)$. 
Then $v = v_1 + v_2$ where $v_1\in V_1$, $v_2 \in V_2$ and $v_1,v_2\in V_3$. Now $v_3$ is a vector space hence $v_1+v_2\in V_3$ and clearly $v_1+v_2\in V_1+V_2$. So $v \in (V_1+V_2)\cap V_3$.
