$\cap\mathfrak{V}$ is a subspace of $V$ Let $V$ be a vector space over $K$ and let $\mathfrak{V}$ be a nonempty set of subspace of $V$. Then $\cap\mathfrak{V}$ is a subspace of $V$.
Proof. Note that $0\in U$ for any $U\in\mathfrak{V}$. Thus, $0\in\cap\mathfrak{V}$. Now, we need to show that $\cap\mathfrak{V}$ is closed under addition and scalar multiplication. But, how? Can you help?

Let $a,b \in \cap \mathfrak{V}$. Then $a,b \in V_i$ for each $V_i \in \mathfrak{V}$. Therefore $a + b \in V_i$ for all $V_i \in \mathfrak{V}$ , hence 
$$a + b \in \cap \mathfrak{V} $$
since the $V_i$ are subspaces of the vectorspace $V$. (Note that this is basically what you did to prove $0 \in \cap \mathfrak{V}$).
Does this gives you an idea for the scalar multiplication?
