A visual proof:A projection $T$ is orthogonal if $||Tx||\leq ||x||$, $x\in V$. Definition A linear transformation $T$ is said to be a projection if $T^2=T$.
A linear transformation $T$ is said to be an orthogonal projection if for each $x\in V$,$||x-Tx||\leq ||x-w||\forall w\in W$. 
Equivalently,if $\langle x-Tx|w\rangle =0 \forall w\in W$. 
We all know that if $T$ is an orthogonal projection of inner product space $V$ on $W$,then $||Tx||\leq ||x||\forall x\in V$.But I want to prove the converse,it seems intuitive to me,but I cannot come up with a formal proof.I am at the same time looking for a good proof which I do not find in any of the answers to the same problem of stack exchange.Till now I think I should proceed like this:

The blue lines indicate the direction of the light,i.e. actually the direction of Kernel $N(T)$.If we take a vector in $V$ perpendicular to the direction of light i.e. we take $0\neq x\in N(T)^{\perp}$.Then the image $Tx$ in $R(T)$ ,I think will satisfy $||Tx||>||x||$,I am saying this by visualization but I could not yet show it formally.Is my intutition correct?How to formally show it?


*

*Also there is a problem,how can I say $N(T)^{\perp}\neq \{0\}$?

 A: 
Proposition.  Let $T$ be an idempotent (projection) map on a (real or complex) Hilbert space $\mathcal{H}$ (that is, $T^2=T$).  Suppose that $\langle\_,\_\rangle$ and $\|\_\|$ are the inner product (antilinear on the right-hand side if the base field is $\mathbb{C}$) and the norm induced by the inner product of $\mathcal{H}$, respectively.  Then, $T$ is an orthogonal projection if and only if  $$\|Tx\|\leq \|x\|$$ for all $x\in\mathcal{H}$.

The forward direction is trivial.  We prove the converse.  We claim that $T$ is self-adjoint, that is,
$$T^\dagger=T\,,$$
where $(\_)^\dagger$ is the Hermitian adjoint operator.  This automatically implies that $T$ is an orthogonal projection.
Since $T\,(T-I)=0$, where $I$ is the identity map on $\mathcal{H}$, we see that 
$$\mathcal{H}=\mathcal{H}_0\oplus\mathcal{H}_1\,,$$
where $\mathcal{H}_\lambda$ is the eigenspace associated with the eigenvalue $\lambda\in\{0,1\}$ of $T$ (because $x=(x-Tx)+(Tx)$ for all $x\in\mathcal{H}$, with $x-Tx\in\mathcal{H}_0$ and $Tx\in\mathcal{H}_1$).  Pick $u\in\mathcal{H}_0$ and $v\in\mathcal{H}_1$.  Observe that $$T(u+v)=Tu+Tv=0+v=v\,,$$ making
$$\|v\|=\big\|T(u+v)\big\|\leq \|u+v\|\,.$$
Therefore,
$$\langle v,v\rangle =\|v\|^2\leq \|u+v\|^2=\langle u+v,u+v\rangle$$
for all $u\in\mathcal{H}_0$ and $v\in\mathcal{H}_1$.  Consequently,
$$\|u\|^2+2\,\text{Re}\big(\langle u,v\rangle\big)=\langle u,u\rangle +\langle u,v\rangle+\langle v,u\rangle \geq 0\,,\tag{*}$$
for all $u\in\mathcal{H}_0$ and $v\in\mathcal{H}_1$.  
If there exists a pair $(a,b)\in\mathcal{H}_0\times\mathcal{H}_1$ such that $\langle a,b\rangle\neq 0$, then we may assume that $\text{Re}\big(\langle a,b\rangle\big)\neq 0$ (if the base field is $\mathbb{R}$, then this is automatically the case, but if the base field is $\mathbb{C}$, then we can replace $a$ by $\text{i}\,a$).  By taking $u:=t\,a$ and $v:=b$ with $t\in\mathbb{R}$ in the inequality (*), we obtain
$$t^2\,\|a\|^2+2\,t\,\text{Re}\big(\langle a,b\rangle\big)\geq 0$$
for every $t\in\mathbb{R}$, but this is a contradiction (by choosing $t:=-\dfrac{\text{Re}\big(\langle a,b\rangle\big)}{\|a\|^2}$, for example).
From the paragraph above, we conclude that $\langle u,v\rangle =0$ for all $u\in\mathcal{H}_0$ and $v\in\mathcal{H}_1$.  Ergo, $\mathcal{H}_0\perp \mathcal{H}_1$.  Now, for $x,y\in\mathcal{H}$, we write $x=x_0+x_1$ and $y=y_0+y_1$, with $x_0,y_0\in\mathcal{H}_0$ and $y_0,y_1\in\mathcal{H}_1$.  Thus, $Tx=x_1$ and $Ty=y_1$, whence
$$\langle Tx,y\rangle =\langle x_1,y\rangle=\langle x_1,y_0+y_1\rangle =\langle x_1,y_1\rangle$$
as $\langle x_1,y_0\rangle=0$, and
$$\langle T^\dagger x,y\rangle=\langle x,Ty\rangle =\langle x_0+x_1,y_1\rangle =\langle x_1,y_1\rangle $$
as $\langle x_0,y_1\rangle=0$.  Consequently,
$$\langle Tx,y\rangle=\langle x_1,y_1\rangle=\langle T^\dagger x,y\rangle$$
for every $x,y\in\mathcal{H}$.  This concludes that $T=T^\dagger$, as asserted.

Theorem.  A projection map $T$ on a Hilbert space $\mathcal{H}$ is an orthogonal projection if and only if $T$ is self-adjoint (i.e., $T^\dagger=T$).
NB.  In this theorem, the definition of orthogonal projections is as given by the OP.  That is, a projection $T:\mathcal{H}\to\mathcal{H}$ such that $$\|x-Tx\|\leq  \|x-w\|$$ for all $w\in \text{im}(T)$.
For the proof, we shall first deal with the converse.  Assume that $T=T^\dagger$.  Let $\mathcal{H}_0$ and $\mathcal{H}_1$ be as before.  We claim that $\mathcal{H}_0\perp \mathcal{H}_1$.  Let $u\in\mathcal{H}_0$ and $v\in\mathcal{H}_1$.  Then, $Tu=0$ and $Tv=v$, whence
$$\langle u,v\rangle =\langle u,Tv\rangle =\langle T^\dagger u,v\rangle=\langle Tu,v\rangle =\langle 0,v\rangle=0\,.$$
Write each $x\in\mathcal{H}$ as $x=x_0+x_1$ with $x_0\in\mathcal{H}_0$ and $x_1\in\mathcal{H}_1$.  Then, $x_1=Tx$, making $x-Tx=x_0$ so $$\|x-Tx\|^2=\|x_0\|^2\,.$$ Observe that 
$$\|x\|^2=\langle x_0+x_1,x_0+x_1\rangle =\|x_0\|^2+\|x_1\|^2\,,$$
as $\langle x_0,x_1\rangle= 0$.  If $w\in\text{im}(T)=\mathcal{H}_1$, then
$$\begin{align}\|x-w\|^2&=\langle x-w,x-w\rangle = \|x\|^2-2\,\text{Re}\big(\langle x,w\rangle\big)+\|w\|^2\\&=\|x_0\|^2+\|x_1\|^2-2\,\text{Re}\big(\langle x,w\rangle\big)+\|w\|^2\,.\end{align}$$
Since $x_0\in\mathcal{H}_0$ and $w\in\mathcal{H}_1$, $\langle x_0,w\rangle=0$, making
$$\langle x,w\rangle =\langle x-x_0,w\rangle=\langle x_1,w\rangle\,.$$
Ergo,
$$\begin{align}\|x-w\|^2&=\|x_0\|^2+\|x_1\|^2-2\,\text{Re}\big(\langle x_1,w\rangle\big)+\|w\|^2\\&=\|x_0\|^2+\|x_1-w\|^2\geq \|x_0\|^2=\|x-Tx\|^2\,.\end{align}$$
This gives the desired inequality $$\|x-Tx\|\leq \|x-w\|\,,$$
for all $x\in\mathcal{H}$ and $w\in \mathcal{H}_1$.  The equality holds if and only if $w=x_1=Tx$.
For the other direction, suppose that $\|x-Tx\|\leq \|x-w\|$ for all $x\in\mathcal{H}$ and $w\in\mathcal{H}_1$.  In particular, this means the inequality holds when $w=0$, that is,
$$\|x-Tx\|\leq \|x\|$$
for all $x\in\mathcal{H}$.  Write $x=x_0+x_1$ with $x_0\in\mathcal{H}_0$ and $x_1\in\mathcal{H}_1$ as before.  We get 
$$\|x_0\|^2=\|x-Tx\|^2\leq \|x_0+x_1\|^2=\|x_0\|^2+2\,\text{Re}\big(\langle x_0,x_1\rangle +\|x_1\|^2\,.$$
This means
$$2\,\text{Re}\big(\langle x_0,x_1\rangle\big) +\|x_1\|^2\geq 0$$
for all $x_0\in\mathcal{H}_0$ and $x_1\in\mathcal{H}_1$.  Use the same argument as the proposition above to verify that $\langle x_0,x_1\rangle =0$.  This means $\mathcal{H}_0\perp \mathcal{H}_1$.  Then, the last paragraph of the proof of the proposition can be used again to show that $T=T^\dagger$.
Corollary.  A projection $T$ on a Hilbert space $\mathcal{H}$ is an orthogonal projection if and only if $$\ker(T)\perp \text{im}(T)\,,$$ that is, for all $x\in\ker(T)$ and $y\in\text{im}(T)$, $\langle x,y\rangle=0$.

Remark.  In case if you are not familar with Hermitian adjoints, for a linear operator $T$ on a Hilbert space $\mathcal{H}$, the Hermitian adjoint of $T$ is defined to be the linear operator $T^\dagger:\mathcal{H}\to\mathcal{H}$ such that
$$\langle T^\dagger x,y\rangle=\langle x,Ty\rangle$$
for all $x,y\in\mathcal{H}$.
