I have a problem proving the following:
Consider a banach space $X$. A projection is bounded if and only if $\mathrm{Im} (P)= \{ Px : x \in X\}$ nd $\mathrm{Ker}(P) = \{x \in X: Px =0\}$ are closed subspaces of $X$.
So we know that a linear operator,say P is a projection if $P^2=P$. Assume first that the projection is bounded. Then $\exists M$ s.t $||Px||\leq M||x||$ for all $x \in X$.
Then, since we know that $X$ is a Banach space, take a sequence say $\{x_n\}_n$ then we know that it is convergent. Say x is the limit. Apply P on $\{x_n\}_n$. Assume $\{Px_n\}$ sonverges to y. Then $x_n - Px_n = (I-P)x_n$ converges to $x - y$. It follows that $y \in \mathrm{Im}(P)$ and $x-y \in \mathrm{Ker}(P)$, so $y = Py = Px$. Hence the conclusion follows from the closed graph theorem.
Is it true what I wrote? I mean does it really proof the theorem? How to proceed next?
There are some things in your proof that need to be investigated.
First of all you do not say that you want to prove the equivalence in a banach space $X$. If you want to then please state it there. If not we cannot use neither the fact that $X$ is a banach space nor the closed graph theorem.
Second. The last paragraph is not clear. It seems that you are assuming that $P$ is continous and then again are proving it.
There are sequences in a banach space that do not converge so this reasoning seems off. This is one thing that you wrote which is not true. Further you do not need to assume that $Px_n$ converges. Since $x_n \to x$ you might follow by using the continuity of $P$ that $Px_n \to Px$.
Here are some hints assuming that we are talking about a banach space $X$. Assume that $P$ is continous. You need to show that both kernel and image of $P$ are closed. Remember what it means for a set $A \subset X$ to be closed. One characterisation is that for all sequences $(x_n)_{n \in \mathbb{N}} \subset A$ converging to an element $x$ in $X$ it holds that $x \in A$.
To show that the kernel is closed consider a sequence $(x_n)_{n \in \mathbb{N}} \subset \mathrm{Ker}(P)$ converging to $x \in X$. This implies that for all $n \in \mathbb{N}$ iut holds that $Px_n = 0$. We need to show that $Px = 0$. Can you take it from here?
So far one doesn't need to use that $P$ is a projection. One can show that the kernel of every bounded linear operator is closed.
Ok, lets talk about the image. Consider a sequence $(y_n) \subset \mathrm{Im}(P) $ converging to $y \in X$. We need to show that $y \in \mathrm{Im}(P)$. This is equivalent to th existence of a $x \in X$ such that $Px = y$. Here in fact you need that $P$ is a projection. We conclude $$ -Py_n = y_n-Py_n +y_n \to y $$ since $y_n - Py_n \in \mathrm{Ker}(P)$. This shows that $$y =\lim\limits_{n \to \infty} P y_n = P (\lim\limits_{n \to \infty} y_n) = Py $$ using the assumption of continuity of $P$.
Another nice way to show this is the fact that $\mathrm{Im}(P) = \mathrm{Ker}(\mathrm{Id}-P)$. Can you show this ? How could one conclude the closedness of $\mathrm{Im}(P)$ using this equality?
The other impilcation of above equivalence is more complicated. Here the closed graph theorem comes in handy. Assume that the mentioned spaces are closed. Consider a sequence $(x_n)_{n \in \mathbb{N}} \subset X$ converging to $x$ such that $Px_n \to y$ given $y \in X $. We need to show that $Px =y$. This is where your arguments work better. Arguing along the lines of your proof we have $x_n -Px_n \to x-y$ this shows $x-y \in \mathrm{Ker}(P)$ since $\mathrm{Ker}(P)$ is closed. Since $\mathrm{Im}(P)$ is closed as well we deduce that there exists $z \in X$ such that $Pz = y$. This leads to $$x-y + 0 = \lim\limits_{n \to \infty} x_n-Px_n + \lim\limits_{n \to \infty} Px_n -z = \lim\limits_{n \to \infty} x_n -z = x-z.$$ This shows that $z = y$ and therefore $Px = P(x-y+y) = 0+ Py = y$. We deduce $Px = y$. The conclusion follows by the closed grph theorem.