If $E$ is a projection with finite-dimensional range $W$ and $||E(a)|| \leq ||a||$ then $E$ is orthogonal 
Prove that if $||E(a)||$ $\leq$ $||a||$ E is a projection with range $W$(finite dimensional) then show that $E$ is orthogonal where $V$ is a real inner product space.

If I can show that $Ker(E)= W^{\perp}$ then I am done.
let $ v \in W^{\perp}$ and $v$ is not in $Ker(E)$ then $E(v) \ne 0$.Now we know that for any projection $$V = R(E) \bigoplus K(E)$$
Then $v = w_1+w_2$ where $w_1 \in R(E)$ and $w_2\in K(E)$.I am trying to show that $w_1 = 0$ and my argument goes like this then
$$||E(w_1+w_2)|| \leq ||w_1+w_2|| \tag{1}$$
$$ \langle w_1+w_2,w_1 \rangle= \langle w_1,w_1 \rangle + \langle w_2,w_1 \rangle=0 \tag{2}$$ $(2)$ holds as $v \in W^{\perp}$.
Then from $(1)$ we can conclude that $E(w_1+w_2)=E(w_1)+E(w_2)= w_1$ then,
$||w_1||$ $\leq$ $||w_1||^2+2 \langle w_1,w_2 \rangle +||w_2||^2= \langle w_1,w_2 \rangle +||w_2||^2$ (from $(2)$)
Now,
$$ \langle w_1+w_2,w_2 \rangle = \langle w_1+w_2,w_2-w_1 \rangle + \langle w_1+w_2,w_1 \rangle = \langle w_1+w_2,w_2-w_1 \rangle =||w_2||^2-||w_1||^2 $$
So, $$||w_1|| \le ||w_2||^2-||w_1||^2 \tag{3}$$
I am stuck in this equation $(3)$.Can someone help me out from here.I know there have been other answers on the same question but I made this approach without consulting the other answers.
Edit 1:With @TheoBendit answer I have been able to show that $ 
W^{\perp} \subset K(E)$. $V$ is infinite dimensional.
To show the opposite that $$Ker(E) \subset W^{\perp}$$
I choose a vector $v \in Ker(E)$ and $ v $ not in $W^{\perp}$ then there is some vector $w \in W^{\perp}$  such that $$\langle v,w \rangle \ne 0$$
Then I try to follow the similar method as given in the answer and I choose the vector $v+nE(w)$
Then $$||E(v+n(E(w))||^2 \le ||v+nE(w)||^2$$
Then $$||nE(w)||^2 \le ||v||^2+||nE(w)||^2+ 2Re\langle(v,E(w))\rangle$$
I wanted to use this to porve that $\langle v,w \rangle =0$.However I cant.
 A: This could have been squeezed into the comment section but I thought it wiser to present it here for added clarity. Assuming a real pre-Hilbert space $\left(V, +, \cdot, \langle \bullet \rangle\right)$ and a projection $p \in \mathrm{Id}\left(\mathrm{End}_{\operatorname{\mathbb{R}-\mathbf{Mod}}}(V)\right)$ (where for any semigroup $(S, \cdot)$ by $\mathrm{Id}(S)$ I mean to symbolise the set of all idempotents of $S$) such that $\lVert p(x)\rVert \leqslant \lVert x \rVert$ for any $x \in V$, the reasoning presented above by Theo Bendit applies to the effect of deducing that $\left(\mathrm{Im} \hspace{1pt} p\right)^{\perp} \subseteq \mathrm{Ker}\hspace{1pt}p$.
The question is how to establish the reverse inclusion and for this we refer to the following proposition:

given a real pre-Hilbert space $V$ and a finite dimensional subspace $U \leqslant_{\mathbb{R}} V$ the decomposition $V=U+U^{\perp}$ holds (it actually yields a so-called internal direct sum decomposition) and furthermore $U=U^{\perp \perp}$.

We omit the proof of this proposition (which is not complicated) and are content to remark that it relies on the fundamental Gram-Schmidt theorem according to which countably dimensional pre-Hilbert spaces have orthonormal bases.
The above proposition is to be applied by noticing that in our current context we on the one hand have $\left(\mathrm{Im} \hspace{1pt} p\right)^{\perp}+\mathrm{Im} \hspace{1pt} p=V$ - since the image of $p$ is finite dimensional by hypothesis - and on the other that $\mathrm{Ker} \hspace{1pt} p \cap \mathrm{Im} \hspace{1pt} p=\{0_V\}$, relations which in conjunction with $\left(\mathrm{Im} \hspace{1pt} p\right)^{\perp} \subseteq \mathrm{Ker}\hspace{1pt}p$ lead straight away to $\left(\mathrm{Im} \hspace{1pt} p\right)^{\perp} = \mathrm{Ker}\hspace{1pt}p$.
P.S. The argument presented by Theo Bendit above can be modified to apply more generally to the same problem where the inner-product is given however not on a real/complex vector space but on a vector space over a real closed field or over its algebraic closure.
A: I know it doesn't really complete your approach, but I came up with a slightly novel approach that I thought I'd share. Suppose $v \in W^\perp$ and $n \in \Bbb{N}$. Then our condition tells us that
$$\|E(v + nE(v))\|^2 \le \|v + nE(v)\|^2.$$
Note that $E(v) \in W$, so $\langle v, E(v)\rangle = 0$, so we can expand the right hand side with Pythagoras' theorem. The left hand side simplifies using linearity and $E^2 = E$. We get
$$(n + 1)^2\|E(v)\|^2 \le \|v\|^2 + n^2\|E(v)\|^2 \implies (2n + 1)\|E(v)\|^2 \le \|v\|^2.$$
Suppose $E(v) \neq 0$. Then, the left hand side tends to $\infty$ as $n \to \infty$, contradicting the above inequality. Thus, $E(v) = 0$, i.e. $v \in W^\perp$.
To get the reverse inclusion, just use dimensions: both $W^\perp$ and $K(E)$ have the same dimension: $\dim V - \dim W$.
A: You may use the elementary fact that if   $\  bt + at^2 \leq 0 \ $ for all $t\in {\Bbb R}$ then $b=0$ and $a\leq 0$.
Let $w\in W=E(V)$ and $z\in W^\perp$. Then:
$$ |w|^2 + 2 t \langle w, Ez\rangle + t^2 |Ez|^2 = |E(w+tz)|^2 \leq |w+tz|^2 =
  |w|^2 + t^2 |z|^2$$
From the above fact, $\langle w,Ez\rangle=0$. As it holds for every $w\in W$ we must have $Ez\in W^\perp$, so finally $Ez\in W\cap W^\perp=\{0\}$.
