Statement: If $V=W_1\oplus W_2$ and $W$ is any subspace such that $W_1\subseteq W$. Show that $W=(W\cap W_1)\oplus (W\cap W_2)$.
Proof: Let $V=W_1\oplus W_2$ and W is any subspace such that $W_1\subseteq W$.
Case 1: $W=W_1$ and $W_1\subseteq W$. Since $w\in W$ iff $w\in W_1$, then if $w\in W_1$ then $w\notin W_2$. Which implies $W\cap W_2=\emptyset$. $\Rightarrow$ $(W_1\cap W)\oplus(W_2\cap W)=W\oplus\emptyset=W$.
Case 2: $W\neq W_1$ and $W_1\subseteq W$. $\Rightarrow$ $\exists w\in W$ such that $w\notin W_1$. However, if $w'\in W_1$ then $w'\in W$. Suppose $\omega\in W$ such that $\omega\notin W_1$. But $W\subset V$, thus $\omega\in V$ which forces $\omega\in W_2$ and implying that $W\cap W_2\neq\emptyset$. Let $U=\{\mu\in V\mid \mu\notin W_1\land\mu\in W_2\}$. Thus, $\omega\in U$, $\forall\omega\in V$ such that $\omega\notin W_1$ and $\omega\in W_2$. Therefore, $W_1\subset W$ and $W\cap W_2=U\neq\emptyset\Rightarrow U\subset W\Rightarrow (W\cap W_1)\oplus(W\cap W_2)=W_1\oplus U\subset W$. Since it is also the case that $W\subset W_1\oplus U$, then $W=(W\cap W_1)\oplus(W\cap W_2)$.
Here is a detailed proof so that you understand the basics. The proof is split in two parts, following the definition of a direct sum: first prove that $\forall w\in W$, there exist $w_1\in W\cap W_1$ and $w_2\in W\cap W_2$, such that $w=w_1+w_2$; second prove that $(W\cap W_1)\cap(W\cap W_2) = \{\mathbf{0}\}$.
$\forall w\in W$, since $w\in V$, there exist $w_1\in W_1$ and $w_2\in W_2$ such that $w=w_1+w_2$. Now since $W_1\subseteq W$, we get $w_1 \in W_1 = W\cap W_1$. Since $W$ is a subspace and $w,w_1\in W$, we get $w_2 = w - w_1 \in W$, so $w_2 \in W\cap W_2$. The first part is thus proved.
The second part follows immediately from the fact that $V = W_1 \oplus W_2$ is a direct sum (so that $W_1\cap W_2=\{\mathbf{0}\}$) and that $\mathbf{0} \in W\cap W_1 \subseteq W_1$, $\mathbf{0} \in W\cap W_2 \subseteq W_2$.