Show that $W\oplus W'=V.$ Let $V$ be an inner product space and $W$ be a subspace of $V$ then show that $W\oplus W'=V.$ Where $W'=\{x\in V|\langle x,y\rangle = 0, \forall y\in W\}.$
My Attempt: It is easy to observe that $W\cap W'=\{0\}.$ Next, I want to show that $V=W+W'.$ I was wondering if I can do the following. Say that for a given vector $v\in V$ find the best approximation $u\in W$ or the projection of $v$ onto $W$ such that $\|u-x\|\leq \|w-x\|, \forall w\in W.$ Then we clearly have that for all $v\in V$ $$v=u+v',v'\in W'.$$ And so $V=W+W'$ Thus we can conclude that $V=W\oplus W'.$ 
Does this work? I am asking this because I read a remark somewhere which mentioned that the best approximation might not always exist if the norm is not induced by the inner product or something like that. So there could be a flaw in my argument. In any case, I would be grateful to get some feedback.  
 A: As noted in the comments, for the best approximation to exist $W$ has to be a closed subspace of $V$.
Other than that, you actually failed to show that $v' = w-v$ is contained in $W'$.
A way to show this is the following lemma:


Let $b, v \in V$ be two vectors. Then $b \perp v$ if and only if $\|b\| \le \|b + \lambda v\|$ for all scalars $\lambda$.

Proof:
$$\|b + \lambda v\|^2 = \|b\|^2 + 2\operatorname{Re} \overline{\lambda}\langle b, v\rangle + |\lambda|^2\|v\|^2$$
If $b \perp v$ then $\|b + \lambda v\|^2 = \|b\|^2 + |\lambda|^2\|v\|^2 \ge \|b\|$.
Conversely, set $\lambda = - \frac{\langle b, v\rangle}{\|v\|^2}$:
$$\|b\|^2 \le \|b + \lambda v\|^2 = \|b\|^2 + 2\operatorname{Re} \overline{\lambda}\langle b, v\rangle + |\lambda|^2\|v\|^2 = \|b\|^2 - \frac{\left|\langle b, v\rangle\right|}{\|v\|^2} \implies \langle b, v\rangle = 0$$

Now for any $v \in V$ let $w \in W$ be the best approximation of $v$ onto $W$. We wish to show $v-w \in W'$, that is $w - v \perp W$.
For any scalar $\lambda$ and $x \in W$ we have:
$$\|v -w - \lambda x \| = \|v - \underbrace{(w + \lambda x)}_{\in W} \| \ge \|v-w\|$$
Therefore $v-w \perp x$. Since $x \in W$ was arbitrary we obtain $v - w \in W'$.
