All projections are continuous in a Banach space. I have this exercise:

A linear operator $P: \Omega \rightarrow \Omega$ is called a
  projection if both the range of P and $P^{-1}(\{0\})$ are closed and
  $P(P(x))=P(x) \forall x$.
Show that if $\Omega$ is a Banach space, then all projections of
  $\Omega$ are continuous.

This exercise is in the chapter of the open mapping theorem, and the closed graph theorem, so it is a pretty big hint that I am supposed to use one of these.
My first attempt (open mapping):
Since $P(\Omega)$ is closed, it is a Banach space. Then if we look at $P: \Omega \rightarrow P(\Omega)$, it is an open mapping. Let U be an open set of $\Omega$, then I need to show that $P^{-1}(U)$, is open. We have that $P^{-1}(U)=P^{-1}(U\cap P(\Omega))$, $U\cap P(\Omega)$ is an open set in $P(\Omega)$. So I need to show that $P^{-1}(U\cap P(\Omega))$ is open.
I don't see how it beeing an open mapping helps me, because that is sadly the other way.
I tried contradiction:
Assume for contradiction that: $P^{-1}(U\cap P(\Omega))$ is not open. Then it must also contain a limit point for its complement.So then there exists x, so that $P(x)\in U$, but $z_n \rightarrow x$, $P(z_n)\ne U$. This implies among other things, that any image of any open ball around x, is both open(open mapping theorem), and contain elements from both U and its complement. But still, this is no contradiction as far as I see.
Any tips?
My second attempt (closed graph theorem).
Since we are working in a Banach space, if I can show that P is a closed operator I will be done.
So assume $\lim x_n = x$, then I must show that if $\lim T(x_n)$ converges, then $\lim T(x_n)=T(x)$.
So assume that $\lim x_n = x$ and that $\lim T(x_n)=y$. Assume for contradiction that $y \ne T(x)$, then we are in the case where $\lim T(x_n)\ne T(x)\rightarrow \lim T(x_n-x)\ne 0$. Hence we are in the case where we have a sequnce $z_n\rightarrow 0$, but $T(z_n)$ converges, but it does not converge to 0. I can't get a contradiction in the case either.
Can you guys please help me?
 A: To complement Ivo's answer using the Closed Graph Thm, here is a proof using the Open Mapping Thm.
Proof. On a Banach space $\left(\Omega,\|\cdot\|_\Omega\right)$ over $\mathbb{R}$ or $\mathbb{C}$, let $P:\Omega \to \Omega$ be a projection, such that both $\text{ran}(P)$ and $\text{Ker}(P)$ are closed subspaces. They are then both Banach spaces under the same norm, and the quotient map $\pi: \Omega \to \Omega/\text{Ker}(P)$ is continuous, where the quotient space is a Banach space under the quotient norm
$\left\|\omega+P(\Omega)\right\|_{\Omega/\text{Ker}(P)}:=\inf\left\{\|\omega'\|_\Omega: \omega'\in \omega+P(\Omega)\right\}\leq \|\omega\|_\Omega.$
Also, since $P^2=P$, the projection map induces an algebraic direct sum, $\Omega=\text{ran}(P)\oplus \text{Ker}(P)$. That is, for every $\omega\in\Omega$, there exist unique elements $x\in\text{ran}(P)$ and $y\in\text{Ker}(P)$ such that
$$\Omega/\text{Ker}(P)\ni\omega+\text{Ker}(P)=x+y+\text{Ker}(P)=x+\text{Ker}(P)\in \pi\left(\text{ran}(P)\right).$$
Hence, the restricted quotient map $\pi|_{\text{ran}(P)}: \text{ran}(P)\to \Omega/\text{Ker}(P)$ is a bounded bijection between Banach spaces. By the Inverse Mapping Thm (Open Mapping Thm), the inverse map $\left(\pi|_{\text{ran}(P)}\right)^{-1}$ is then also bounded. Therefore, the projection map is a composition of two continuous maps: $P=\left(\pi|_{\text{ran}(P)}\right)^{-1}\circ \pi$. ${\rm\square}$
A: Let's use the closed graph theorem. We want to prove that: $${\rm gr}(P) = \{ (x,P(x)) \in \Omega \times \Omega \mid x \in \Omega  \}$$is closed. Let $((x_n, P(x_n)))_{n \geq 1}$ be a sequence in the graph such that $x_n \to x$ and $P(x_n) \to y$. Since $(P(x_n))_{n \geq 1}$ is a sequence in the range of $P$, which we assume closed, we have that $y$ is in the range of $P$, so we write $y = P(\tilde{x})$ for some $\tilde{x} \in \Omega$. 
Note that $x_n - P(x_n) \to x - y$, but $(x_n-P(x_n))_{n \geq 1}$ is a sequence in $P^{-1}(\{0\})$ because of the $P^{\circ 2} = P$ property. Since $P^{-1}(\{0\})$ is closed, we have $x-y \in P^{-1}(\{0\})$, so $P(x) = P(y)$. Well, using $P^{\circ 2} = P$ again we have: $$P(\tilde{x}) = y \implies P(x) = P(y) = P(\tilde{x}) = y.$$So the graph is closed, and by the closed graph theorem, $P$ is continuous.
(I really enjoyed solving this, thanks for the opportunity)
