Prove that the map $E:V\to V$ is linear. Let $V$ be an inner product space and $W$ be a finite dimensional subspace of $V.$ Define $E:V\to V$ such that $$E(x)=u$$ where $u$ is the unique best approximation of $x$ on $W$ or in other words $$||x-u||\leq ||x-w||\forall w\in W.$$
I want to show that $E$ is linear. Now my stratergy was to first consider $\gamma\in \mathbb{K}$ and let $E(x)=u.$ We want to show that $E(\gamma x)=\gamma u.$ Since $$||x-u||\leq ||x-w||$$ for any $w\in W$ we have that $$|\gamma||x-u||\leq |\gamma| ||x-w||\implies ||\gamma x-\gamma u||\leq ||\gamma x-\gamma w ||=||\gamma x-w'||.$$
Can I conclude from this that $E(\gamma x)=\gamma u?$
Next, I tried showing that $E(x+y)=E(x)+E(y).$ So first let $E(x)=u$ and $E(y)=u'.$ Then we want to show that $E(x+y)=u+u'.$
Consider $$||x+y-(u+u')||\leq ||x-u||+||y-u'||\leq ||x-w||+||y-w||\forall w\in W.$$
How do I proceed? 
 A: I don't know how to prove that $E$ is linear along these lines. Here's a way of doing it (if you are not interested in it, I will delete this answer): take an orthonormal basis $\{e_1,e_2,\ldots,e_n\}$ of $W$ and prove that$$(\forall v\in V):E(v)=\sum_{k=1}^n\langle v,e_k\rangle e_k.$$It is clear then that $E$ is linear.
A: If you know or can prove that $x-u$ is orthogonal to $W$ (using the initial notations of your post), then you’re done as this proves that $u$ is the orthogonal projection of $x$ on $W$. And the orthogonal projection is a linear map.
A: First we look for the best approximation of $0$:
$$
E(0) = u_0
$$
would give
$$
\lVert 0 - u_0 \rVert\le \lVert 0 - w\rVert \quad (w \in W)
$$
As $W$ is a subspace of $V$ we have $0 \in W$ and thus
$$
\lVert 0-u_0\rVert = \lVert u_0 \rVert \le \lVert 0 - 0 \rVert = 0
$$
and $u_0 = 0$ and $E(0) = 0$ follows.
Then
$$
E(\gamma x) = u_{\gamma x}
$$
with
$$
\lVert \gamma x - u_{\gamma x} \rVert \le \lVert\gamma x - w \rVert \quad (w \in W)
$$
and for $\gamma \ne 0$:
$$
\lvert \gamma \rvert \lVert x - u_{\gamma x}/\gamma \rVert
\le
\lvert \gamma \rvert \lVert x - w/\gamma \rVert \quad (w \in W) \iff \\
\lVert x - u_{\gamma x}/\gamma \rVert
\le
\lVert x - w' \rVert \quad (w' \in W)
$$
as $\{w/\gamma \mid w \in W\} = W$, so we see that due to the uniqueness of the best approximation $u_{\gamma x}/\gamma = u$ and
$$
E(x) = u = u_{\gamma x} / \gamma = E(\gamma x) / \gamma \iff \\
E(\gamma x) = \gamma E(x)
$$
This stays true, if we include $\gamma = 0$, as initially shown.
