If $V_1$ and $V_2$ are subspaces of vector space $V$, where $V$ is a finite-dimensional inner product space then:
$(V_1+V_2)^⊥=V_1^⊥+V_2^⊥$
So far I have tried showing this, by taking a vector $u \in V_1^⊥+V_2^⊥ $
Then $u=k+w $ with $k\in V_1^⊥$ and $ w \in V_2^⊥$, so if I take a vector $v \in V_1+V_2 $ then I have to show that $ <u,v>=0$. I can also write $v=a+b$ with $a\in V_1$ and $b\in V_2$ but at this point I'm stuck.
This statement is not true. For example, take $V=\Bbb{R}^2$, $V_1=\langle(1,0)^t\rangle$, and $V_2=\langle(0,1)^t\rangle$. Now, we have $V_1^\bot=\langle(0,1)^t\rangle$ and $V_2^\bot=\langle(1,0)^t,(0,1)^t\rangle$, so:
$$V_1^\bot+V_2^\bot=\langle(0,1)^t,(1,0)^t\rangle=\Bbb{R}^2$$
However, $V_1+V_2=\langle(1,0)^t,(0,1)^t\rangle$, meaning: $$(V_1+V_2)^\bot=\langle(0,0)^t\rangle$$
Now, according to this question, we actually have the following: $$(V_1+V_2)^\bot=V_1^\bot \cap V_2^\bot$$ $$(V_1 \cap V_2)^\bot=V_1^\bot + V_2^\bot$$
