let $v \in V$ such that $W=[v]$. It is desired to show that $$\|v-E(v)\| \leq \|v-w\|,\;\;\;\; \forall v \in V,\; and\;\;\forall w \in W$$ Where $E$ is the orthogonal projection of V in W.
Here's what I did: I simply used Cauchy-Schawrz inequality and the fact that $E(v)=w$. then, $$\|v-w\|^2={\langle v-w, v-w \rangle} \leq \|v-w\|^2$$ but $E(v)=w$. therefore, $${\langle v-E(v),v-E(v) \rangle} \leq \|v-w\|^2\\ \implies\|v-E(v)\|^2 \leq \|v-w\|^2 $$ Which lead us to the desired answer. My question is: is it right? because since i assumed $E(v)=w$ wouldn't it imply that instead of an inequality sign I should have an "equal" sign? Any tips are welcome.
No, it is not right. You did not prove that you have the inequality $\|v-E(v)\|\leqslant\|v-w\|$ for any $v\in V$ and any $w\in W$; you only proved that $\|v-E(v)\|\leqslant\|v-E(v)\|$ for any $v\in W$, which is trivial (you always have $x\leqslant x$). Besides, do you really need the Cauchy-Schwarz inequality in order to deduce that $\|v-w\|\leqslant\|v-w\|$?
You can prove it noting that\begin{align}\|v-w\|^2&=\|v-E(v)+E(v)-w\|^2\\&=\|v-E(v)\|^2+2\langle v-E(v),E(v)-w\rangle+\|E(v)-w\|^2\\&=\|v-E(v)\|^2+\|E(v)-w\|^2 \text{ ($v-E(v)$ and $E(v)-w$ are orthogonal)}\\&\geqslant\|v-E(v)\|^2.\end{align}