Suppose that $W_1$ and $W_2$ are both four-dimensional subspaces of a vector space $V$ of dimension seven. Explain why $W_1 \cap W_2 \neq \{0\}$. 
Suppose that $W_1$ and $W_2$ are both four-dimensional subspaces of a vector space $V$ of dimension seven.
  Explain why $W_1 \cap W_2 \neq \{0\}$.

Suppose $W_1\cap W_2 = \{0\}$,
since $\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1\cap W_2)$ and $\dim(W_1)=\dim(W_2)=4$, $\dim(V)=7$,
$$\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(\{0\})=4+4-0= 8 > \dim(V)=7$$
which does not make sense since both $W_1$, $W_2$ are subspaces of $V$.
Therefore, $W_1 \cap W_2 \neq \{0\}$
This is how I solved it. Is it right?

Your proof is fine, but there's no need to use contradiction. 
By Grassmann’s formula
$$
7=\dim V\ge\dim(W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)=
4+4-\dim(W_1\cap W_2)
$$
Therefore
$$
\dim(W_1\cap W_2)\ge 4+4-7=1
$$
