Note that $\|Tx\|=\|x\|$ is the definition of an isometry. Over a finite-dimensional real inner product space, this is equivalent to the matrix of $T$ in an orthonormal basis being "orthogonal", i.e. $A^TA=AA^T=I_n$.
Orthogonal projection means $T^2=T$ and $T^*=T$, i.e. self-adjoint idempotent. The first thing you should remark is that this is equivalent to $T$ being idempotent with $\ker T \perp \mbox{im} T$. Indeed, if $T$ is idempotent, then $T^*$ is idempotent with $\ker T^*=(\mbox{im} T)^\perp$ and $\mbox{im} T^*=(\ker T)^\perp$.
Claim: if $T$ is idempotent and $\|T\|\leq 1$ (i.e. $\|Tx\|\leq \|x\|$ for all $x\in V$), then $T$ is self-adjoint (i.e. $T$ is an orthogonal projection).
Remark: the converse is true since then the orthogonal direct sum $V=\ker T\oplus \mbox{im} T$ yields $\|x+y\|^2=\|x\|^2+\|y\|^2\geq \|y\|^2=\|0+Ty\|^2=\|T(x+y)\|^2$ for all $x\in \ker T$ and all $y\in \mbox{im} T$. Note that all this holds on a general inner product space. No need to assume finite dimension.
Proof: we need to prove that $(x,y)=0$ for every $x\in \ker T$ and every $y\in\mbox{im} T$. Let us take two such vectors, which are characterized by $Tx=0$ and $Ty=y$. Then for every $t\in\mathbb{R}$
$$
t^2\|y\|^2=\|ty\|^2=\|T(x+ty)\|^2\leq \|x+ty\|^2=\|x\|^2+2t\,\mbox{Re}(x,y)+t^2\|y\|^2
$$
whence
$$
\mbox{Re}(x,y)\geq -\frac{\|x\|^2}{2t}\;\forall t>0\qquad\mbox{and}\qquad \mbox{Re}(x,y)\leq -\frac{\|x\|^2}{2t}\;\forall t<0
$$
which implies $\mbox{Re}(x,y)=0$ by letting $t$ tend to $\pm \infty$. In the real case, we are done since $\mbox{Re}(x,y)=(x,y)=0$. In the complex case, take $e^{i\theta}$ such that $|(x,y)|=e^{i\theta}\mbox{Re}(x,y)=\mbox{Re}(x,e^{i\theta}y)$ and apply the above to $x,e^{i\theta}y$ to conclude that $|(x,y)|=0$ whence $(x,y)=0$. QED.
Ed. Here is another proof which would prove directly $\operatorname{dim}(U∩W^{\perp})=\operatorname{dim}(W∩U^{\perp})$ if the dimension $n$ is a finite number. (You statement automatically follows)
Let $T=U\cap W^{\perp},V=W\cap U^{\perp}$, and decompose $U^{\perp}$ and $W$ as $U^{\perp}=V \oplus U_1$, $W=V \oplus V_1$. It's not hard to see that $U^{\perp}+W=U_1\oplus V \oplus W_1$, and $\operatorname{dim}(U^{\perp}+W)=n-\operatorname{dim}(V)$. Clearly $T$ is othogonal to $U^{\perp}+W$, so $U^{\perp}+W+T=(U^{\perp}+W)\oplus T$ is a subspace of $n$-dimensional space, so $\operatorname{dim}(U^{\perp}+W)+\operatorname{dim}(T) \le n$, or $n-\operatorname{dim}(V)+\operatorname{dim}(T) \le n$ $\implies$ $\operatorname{dim}(T) \le \operatorname{dim}(V)$. You can similarly prove that $\operatorname{dim}(T) \ge \operatorname{dim}(V)$. We therefore have $\operatorname{dim}(T) = \operatorname{dim}(V)$.
Below was my first/original proof, more an "elementary school style".
First, the following is true: if the intersection of two subspaces $X$ and $Y$ is $\{0\}$, assume ${a_1,...,a_m}$ and ${b_1,...,b_l}$ are the bases of $X$ and $Y$, then ${a_1,...,a_m,b_1,...,b_l}$ are linearly independent, i.e. if a linear combination of vectors ${a_1,...,a_m,b_1,...,b_l}$ is zero, then all coefficients must be zero. This implies that $\operatorname{dim}(X+Y)=\operatorname{dim}(X)+\operatorname{dim}(Y)$. Its proof should be trivial.
Second, if the statement is false, i.e. $W∩U^{\perp}=\{0\}$,then $\operatorname{dim}(W+ U^{\perp})=\operatorname{dim}(W)+\operatorname{dim}(U^{\perp})=\operatorname{dim}(U)+\operatorname{dim}(U^{\perp})=n$, so the bases of $W$ and $U^{\perp}$ should be linearly independent, and any vector can be expressed as the sum of a vector in W and a vector in $U^{\perp}$.
However, we know that $U$ contains a nonzero vector that is in $W^{\perp}$, say $z \in W^{\perp}$. We then can find $x \in W$ and $y \in U^{\perp}$ such that $z=x+y$. Let's check the inner product of $z$ with itself. $(z,z)=(x+y,z)=(x,z)+(y,z)$. We know that $(x,z)=0$ because $z \in W^⊥$ and $(y,z)=0$ because $z \in U$. Therefore $(z,z)=0$, which is clearly wrong.
Therefore $W$ contains at least one nonzero vector that is in $U^{\perp}$.
We can also prove that $\operatorname{dim}(U∩W^{\perp})=\operatorname{dim}(W∩U^{\perp})$ if the dimension $n$ is a finite number.
Best Answer
By the dimension theorem, $\dim V=\dim N(T) +\dim R(T)$, so whenever these have nontrivial intersection (e.g. with $T(x, y) =(0, x)$) we don't have $V=N(T) +R(T) $.
A linear operator $T$ is a projection iff it is idempotent, i.e. $T^2 =T$.
Then any vector $x$ can be decomposed as $x=(x-Tx) \ +\ Tx\ \in N(T) +R(T)$, and if $x\in N(T) \cap R(T)$, then $Tx=0$ and $x=Ty$ for some $y$, so $x=Ty=TTy=Tx=0$.
It shows that in this case we indeed have $V=N(T)\oplus R(T)$, and $T$ effectively projects $a+b\mapsto b$.
Conversely, if $V=A\oplus B$ then the projection $a+b\ \mapsto b$ is clearly idempotent.
To talk about orthogonality, an inner product has to be considered on $V$, which induces an adjoint $T^*$ for every linear operator $T$, uniquely determined by the equation $\langle Tx, y\rangle =\langle x, T^*y\rangle$ (in an orthonormal basis, the matrix of $T^*$ is just the (complex conjugated) transpose of the matrix of $T$).
Now, if $T^2=T$, we will have $N(T) \perp R(T)$ iff $T^*=T$ (selfadjoint).
Supposed $T^*=T=T^2$, if $a\in N(T) $ and $b\in R(T) $, we have $$\langle a, b\rangle =\langle a, Tb\rangle =\langle a, T^*b\rangle =\langle Ta, b\rangle = 0$$ meaning $a\perp b$.
Conversely, if $A\perp B, \ V=A\oplus B$ and $T=a+b\mapsto b$ is an orthogonal projection, then, with the decompositions $x=x_A+x_B, \ y=y_A+y_B$, we have $$\langle Tx, y\rangle = \langle x_B, \, y_A+y_B\rangle =\langle x_B, y_B\rangle=\langle x, Ty\rangle$$ showing $T^*=T$.