I think I am misunderstanding something about this exercise (and about adjoints in general).
Exercise: Suppose $P \in L(V)$ is such that $P^2 = P$. Prove that if $P$ is an orthogonal projection then $P$ is self-adjoint.
Answer: Because $P$ is an orthogonal projection, there exists a subspace $U$ of $V$. Let's decompose $v = u + w$ where $u \in U$ and $w \in U^\bot$.
$ \langle P^2v, u\rangle = \langle Pv, u\rangle = \langle u, u\rangle$ where the first equality comes from $P^2 = P$ and the second one because $P$ is the orthogonal projection.
Also, $\langle P^2v, u\rangle = \langle Pv, P^*u\rangle=\langle u, P^*u\rangle$ where the first equality comes from the adjoint and the second one because $P$ is the orthogonal projection.
That leaves us with $\langle u, u\rangle=\langle u, P^*u\rangle$ and by uniqueness of $P^*u$ then $u=Pu=P^*u$ so $P=P^*$.
Question: Am I allowed to pick $v$ and $u$ the way I did? I feel I am missing something.
Thanks as always!
Edit: I was messing up with the definition of adjoint. The "uniqueness" part made no sense. Robbie answer is good.
Best Answer
I'm not sure what you mean by 'uniqueness of $P^\ast u$'. In any case you have to pick a generic vector for the proof – i.e. you need to show $$ \langle Px, y \rangle = \langle x, Py \rangle $$ for any vectors $x,y$. You can't pick $x = Py$.
Note that orthogonal projection is a stronger condition than $P^2 = P$ so, intuitively, you won't need to use the former in the proof, but you will want to use the fact that the spaces are orthogonal, which I don't think you have used.
For clarity I'm going to change notation slightly: write $u = u_1 + u_2$, $v = v_1 + v_2$ where $u_i \in U_i$ and $P(V) = U_1$. Then $$ \langle Pu, v \rangle = \langle u_1, v_1 + v_2 \rangle = \langle u_1, v_1 \rangle =\cdots= \langle u, Pv \rangle $$ Can you now see how to complete the proof?