Range of linear operator $P$ and its adjoint $P^*$ 
Let $\mathcal{H}$ be a Hilbert space, $P:\mathcal{H}\to\mathcal{H}$ a bounded linear operator, $P^*$ be the adjoint, and $R(P),N(P)$ be the range and nullspace.
  (a) Show that $N(P^*) = R(P)^\perp$ and $\overline{R(P^*)} = N(P)^\perp$
  (b) Show that $R(P^*)$ is closed if $R(P)$ is closed
  (c) Assume that $P^2 = P$.  Show that the following are equivalent
  $\quad$ (i) $R(P)$ is closed and $||x-Px|| = \inf_{y\in R(P)}||x-y||$ for every $x\in \mathcal{H}$
  $\quad$ (ii) $P=P^*$

$(a)$ is an exercise I've done before, but for the rest I haven't been able to make progress.
For $(b)$, if I directly apply $(a)$, then I obtain the closure $\overline{R(P^*)} =  N(P)^\perp$, which is not useful.  
 A: Solution for (b):
Lemma $1$:

Let $H$ and $K$ be Hilbert spaces and $T : H \to K$ bounded linear map. If $T$ is surjective, then $T^*$ is bounded from below.

Lemma $2$:

Let $X$ and $Y$ be Banach spaces and $T : X \to Y$ linear map bounded from below. Then $R(T)$ is closed.

Assume $R(P)$ is closed. Consider the corestriction $P_0 : H \to R(P)$ given by $P_0x = Px, \forall x \in H$. $P_0$ is surjective and $R(P)$ is a Hilbert space so by Lemma $1$, the adjoint $P_0^* : R(P) \to H$ is bounded from below. Lemma $2$ gives that $R(P_0^*)$ is closed in $H$. We claim $R(P_0^*) = R(P^*)$. Clearly $R(P_0^*) \subseteq R(P^*)$. The converse follows from
$$R(P^*) \subseteq  \overline{R(P^*)} =  N(P)^\perp = N(P_0)^\perp = R(P_0^*)$$
Hence $R(P^*)$ is closed in $H$. The converse implication follows from $P^{**} = P$.

Solution for (c):

The condition $P^2 = P$ implies $Py = y \iff y \in R(P)$.
Namely, if $y = Py$ then clearly $y \in R(P)$. Conversely, if $y \in R(P)$ then $\exists x \in H$ such that $Px = y$. We have
$$Py = P^2x = Px = y$$
Now notice that $H = R(P) \,\dot+\, N(P)$. The intersection is trivial since $y \in R(P) \cap N(P)$ implies $y = Py = 0$. Furthermore, any $y \in H$ can be expressed as $$y = \underbrace{Py}_{\in R(P)} + \underbrace{y - Py}_{\in N(P)}$$
The range $R(P)$ is always closed. Indeed, take a sequence $(y_n)_n$ in $R(P)$ such that $y_n \xrightarrow{n\to\infty} z \in H$. Applying the bounded map $P$ we get
$$y_n = Py_n \xrightarrow{n\to\infty} Pz$$
but $y_n \xrightarrow{n\to\infty} z$ so $Pz = z$ implying $z \in R(P)$.
Lemma $3$ (Birkhoff-James):

Let $b, v \in H$. We have
$$b \perp v \iff \|b\| \le \|b + \lambda v\|, \forall \lambda \in \mathbb{F}$$

Assume $\|x - Px\| = \inf_{y \in R(P)} \|x-y\|$ for all $x \in H$. We claim $x - Px \perp Px$. We have:
$$\|x-Px + \lambda Px\| = \|x - \underbrace{(\lambda-1)Px}_{\in R(P)}\| \ge \|x - Px\|, \forall \lambda \in \mathbb{F}$$
Therefore $x - Px \perp Px$, by Lemma $3$. Take $u \in R(P)$ and $v \in N(P)$. Considering $x = u + v$ and the previous relation gives $u \perp v$. We conclude $R(P) \perp N(P)$ so
$$H = R(P) \oplus N(P)$$
Finally
$$\langle Px, y\rangle = \langle Px, y - \underbrace{(y - Py)}_{\perp Px}\rangle= \langle Px, Py\rangle = \langle Px + \underbrace{(x - Px)}_{\perp Py}, Py \rangle = \langle x, Py\rangle, \forall x,y \in H$$
so $P^* = P$.

Conversely, assume $P^* = P$. Then $R(P) = R(P^*)$ so
$$H = \overline{R(P^*)} \oplus N(P) = R(P) \oplus N(P)$$

In particular $x - Px \perp P(x-y)$ for all $x \in H$ and $y \in R(P)$.
Therefore, by Lemma 3:
$$\inf_{z \in R(P)}\|x-z\| \le \|x-Px\| \le \|x - Px + P(x - y)\| = \|x - Py\| = \|x-y\|$$
Taking the infimum over $y \in R(P)$ gives $\|x-Px\| = \inf_{y \in R(P)}\|x-y\|$.
I'll write the proofs for the lemmas if needed.
A: Here are my thoughts for part (c). 
First note that if $P$ is a bounded linear operator such that $P^2=P$, then $R(P)$ is closed. This is because $R(P)=N(I-P)$ and $I-P$ is continuous. Furthermore, we can decompose $H$ as $R(P)\oplus N(P)$. This is because if $x \in H$, then $x=Px+(x-Px)$ and $Px \in R(P)$ while $P(x-Px)=Px-Px=0$ and hence $x-Px \in N(P)$. Also $x \in R(P) \cap N(P)$ implies that $Px=0$ and $x=Py$ for some $y$. Hence $x=Py=P^2y=Px=0$.
We are now ready to prove the equivalence in part (c). Suppose that $R(P)$ is closed and $\Vert x-Px\Vert = \inf_{y \in R(P)}\Vert x-y\Vert$. This implies that $P$ is the orthogonal projection of $H$ onto $R(P)$ since the above expression uniquely defines the orthogonal projection. This implies that $N(P)=R(P)^\perp$. We will now see that $P$ is self-adjoint. Let $x,y \in H$ be given. Let us write $x=x_1+x_2$ and $y=y_1+y_2$ where $x_1,y_1\in R(P)$ and $x_2,y_2 \in N(P)=R(P)^\perp$. It follows that 
$$(Px,y)-(x,Py) = (Px_1+Px_2,y_1+y_2)-(x_1+x_2,Py_1+Py_2)$$
Since $Px_2=Py_2=0$, we have that this is equal to
$$(Px_1,y_1)+(Px_1,y_2)-(x_1,Py_1)+(x_2,Py_1) $$
Since $y_2,x_2 \in R(P)^\perp$ and $Px_1,Px_2 \in R(P)$ we have that this simplifies to 
$$(Px_1,y_1)-(x_1,Py_1)=(x_1,y_1)-(x_1,y_1)=0 $$
as $x_1,y_1 \in R(P)$ implies that $Px_1=x_1$ and $Py_1=y_1$. Thus we have that $(Px,y)=(x,Py)$ for all $x,y \in H$ and thus $P$ is self-adjoint.
Now conversely suppose that $P=P^\ast$. From the first point, we know that $R(P)$ is closed. Thus we just have to show that, for all $x \in H$, $\Vert x-Px \Vert = \inf_{y \in R(P)} \Vert x-y\Vert$. Let $x \in H$ and $y \in R(P)$ be given. Since $y \in R(P)$, we have that $Py=y$. It then follows that
$$(x-Px,Py)=(x,Py)-(Px,Py)=(x,Py)-(Px,y)=(x,Py)-(x,P^\ast y)=(x,Py)-(x,Py)=0$$
Thus by Pythagoras's theorem 
$$\Vert y-x\Vert^2=\Vert y-Px+Px-x\Vert^2=\Vert y - Px\Vert ^2 + \Vert Px-x\Vert^2 \ge \Vert Px-x\Vert^2 $$
Thus we have that $\Vert Px - x \Vert \le \inf_{y \in R(P)} \Vert y-x\Vert$. Since $Px \in R(P)$, we have that the above are in fact equal as required.
