If $T$ is a projection, then $R(T)=N(T)^\perp$ if and only if $N(T)=R(T)^\perp$. In Friedberg's Linear Algebra textbook, a linear operator $T$ on an inner product space $V$ is said to be an orthogonal projection if $T$ is a projection, and
$$\tag{1}R(T)=N(T)^\perp$$
and
$$\tag{2}N(T)=R(T)^\perp\mbox{,}$$ where $S^\perp$ denotes the orthogonal complement of $S$. It is shown at once that when $V$ is finite-dimensional, (1) is equivalent to (2).
I think that the equivalence is valid even if $V$ is infinite-dimensional. For suppose (1) holds. It suffices to show that $N(T)=N(T)^\perp\left.\right.^\perp$. That $N(T)\subset N(T)^\perp\left.\right.^\perp$ is evident. Since $T$ is a projection, $V=R(T)\oplus N(T)$, hence every vector $v\in V$ can be expressed uniquely as a sum of $v_1\in R(T)$ and $v_2\in N(T)$. If $v\not\in N(T)$, then $v_1\not=0$, thus
$$\left<v,v_1\right>=\lVert v_1\rVert^2>0\mbox{.}$$
This together with (1) shows that $v\not\in N(T)^\perp\left.\right.^\perp$; therefore $N(T)^\perp\left.\right.^\perp\subset N(T)$. Similarly, (2) implies (1).
Am I right?
 A: I think your approach is correct. Namely, since $T^2 = T$ it easily follows $$V =R(T) \dotplus N(T)$$
and if you assume $(1)$, then $R(T) \perp N(T)$ so the  sum is in fact orthogonal:
$$V =R(T) \oplus N(T).$$
In particular, both subspaces $N(T)$ and $R(T)$ are closed and hence $(2)$ follows.
Moreover, $T$ in fact must be bounded. Namely, for any $x \in V$ we have $Tx \perp x - Tx$ so
$$\|Tx\|^2 \le \|Tx\|^2 + \|x-Tx\|^2 = \|Tx+(x-Tx)\|^2 = \|x\|^2.$$
Note: The following example is wrong as the hermitian form $\langle \cdot,\cdot\rangle$ defined below is not an inner product.
Consider $c_{00}$, the space of all finitely-supported sequences in $\Bbb{C}$ equipped with the (incomplete) inner product
$$\langle (x_n)_n,(y_n)_n\rangle := \sum_{n=1}^\infty\left(x_n\overline{y_n} + nx_{2n}\overline{y_{2n-1}} + nx_{2n-1}\overline{y_{2n}}\right).$$
Consider
$$T : c_{00} \to c_{00}, \quad T(x_n)_n := \left(x_1+x_2, 0, x_3+2x_4, 0, x_5+3x_6, 0,\ldots\right).$$
Then $T^2 = T$ and
$$R(T) = \operatorname{span}\{e_{2n-1} : n \in \Bbb{N}\}, \quad N(T) = \operatorname{span}\{e_{2n}-ne_{2n-1} : n \in \Bbb{N}\}$$
and we have the orthogonal sum
$$c_{00} = R(T) \oplus N(T).$$
However, $Te_{2n} = ne_{2n-1}$ for all $n \in \Bbb{N}$ so $T$ is unbounded.
