You can also directly verify this by definition of self-adjointness:

Let $U$ be a subspace of $V$, let $P$ be the orthogonal projection that projects onto $U$; let $x, y \in V$, and we can decompose them as $x=u_x+v_x, y=u_y+v_y$, with $u_x, u_y \in U, v_x, v_y \in U^\perp$, so $\langle u_x, v_y \rangle = \langle v_x, u_y \rangle = 0 $. By definition of $P$, $P x = u_x, P y = u_y$.

Then $$ \langle Px, y \rangle = \langle u_x,u_y+v_y \rangle = \langle u_x,u_y \rangle = \langle u_x + v_x, u_y \rangle = \langle x, Py \rangle $$ it follows $P$ is self-adjoint and its matrix is Hermitian (i.e. symmetric when $V$ is real).

You can also directly verify this by definition of self-adjointness:

Let $U$ be a subspace of $V$, let $P$ be the orthogonal projection that projects onto $U$; let $x, y \in V$, and we can decompose them as $x=u_x+v_x, y=u_y+v_y$, with $u_x, u_y \in U, v_x, v_y \in U^\perp$, so $\langle u_x, v_y \rangle = \langle v_x, u_y \rangle = 0 $. By definition of $P$, $P x = u_x, P y = u_y$.

Then $$ \langle Px, y \rangle = \langle u_x,u_y+v_y \rangle = \langle u_x,u_y \rangle = \langle u_x + v_x, u_y \rangle = \langle x, Py \rangle $$ it follows $P$ is self-adjoint and its matrix is Hermitian (i.e. symmetric when $V$ is real).

UPDATE: I just read your question again; here's a more intuitive explanation.

Remember:

1. $\forall A \in \mathbb{R}^{n\times n} $, $\operatorname{null}(A)$ and $\operatorname{row}(A)$ are always orthogonal complements of each other in $\mathbb{R}^n$
2. when $A$ describes a projection, $A^2=A$, $\mathbb{R}^n = \operatorname{null}(A) \oplus \operatorname{col}(A)$

Now when $A$ describes an orthogonal projection, we also have $\operatorname{null}(A) \perp \operatorname{col}(A)$, so $\operatorname{null}(A)$ and $\operatorname{col}(A)$ are orthogonal complements of each other in $\mathbb{R}^n$; this happens iff $A$ is symmetric, so that $\operatorname{row}(A) = \operatorname{col}(A)$.