Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp $ 
Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $$(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp $$

My Try
One direction is easy : Let $\alpha \neq 0$ and $\alpha \in W_1^\perp + W_2^\perp$, i.e. $\alpha$ can be written as $\alpha = \beta + \gamma$ such that $\beta \in W_1^\perp$ and $\gamma \in W_2^\perp$, hence $(\beta|\eta)=0$ for all $\eta \in W_1 $ and $(\gamma|\delta)=0$ for all $\delta \in W_2 $. Now for all $\eta \in W_1 \cap W_2$ it is clear that $(\alpha | \eta) =0$ . hence 
$$(W_1 \cap W_2)^\perp\supset W_1^\perp + W_2^\perp $$
For proving the other containment, let $\alpha \neq 0$ and $\alpha \in (W_1 \cap W_2)^\perp$, it means that for all $\beta \in W_1 \cap W_2$, $(\alpha|\beta)=0$. Hence $\alpha \in V \setminus (W_1 \cap W_2) = V \setminus (W_1) \cup  V \setminus  (W_2)$. Hence $\alpha \in W_1^c$ or $\alpha \in W_2^c$. WLOG suppose $\alpha \in W_1^c$. We also have that $$V=(W_1 ) \oplus (W_1 )^\perp$$therefore $\alpha = \eta + \delta $ where $\eta \in W_1$ and $\delta \in W_1^\perp$ and indeed $\delta \neq 0$. From here I have to somehow show that $\eta=0$, but I am stuck in here...
I already appreciate any help
 A: It is not hard to see that $(W_1+W_2)^\perp=W_1^\perp\cap W_2^\perp$. Try to prove that if you haven't already. A proof can be found here.
Since the spaces are finite dimensional, $(W_i^\perp)^\perp=W_i$ for $i=1,2$. Then using these two facts, observe
$$
(W_1\cap W_2)^\perp=((W_1^\perp)^\perp\cap (W_2^\perp)^\perp)^\perp=((W_1^\perp+W_2^\perp)^\perp)^\perp=W_1^\perp+W_2^\perp.
$$
A: Some relevant facts are (1)  that if $W$ is a closed subspace, then $W^{\bot \bot} = W$, and (2) if $W \subset X$, then $X^\bot \subset W^\bot$, and (3)
finite dimensional subspaces are always closed (and so (1) applies).
You have shown one direction, you wish to show that
$(W_1 \cap W_2)^\bot\subset W_1^\bot + W_2^\bot$.
Because of the above facts, this is equivalent to showing $W_1 \cap W_2 \supset (W_1^\bot + W_2^\bot)^\bot$.
So, suppose that $x \in (W_1^\bot + W_2^\bot)^\bot$. This means that
$\langle x,w_1'+w_2'\rangle = 0$ whenever $w_k' \in W_k^\bot$, $k=1,2$.
In particular, we have $\langle x,w_1'\rangle = 0$ for all $w_1' \in W_1^\bot$, and so
$x \in W_1^{\bot\bot} = W_1$. Similarly, we have $x \in W_2$. And so, $x \in W_1 \cap W_2$.
