I have seen similar questions on this site, most notably this one: Convergence of projections onto a nested sequence of subspaces of a Hilbert space, but they all include the Hilbert space assumption. The following, from Cheney's Analysis for Applied Mathematics, is more general:
Let $P_1, P_2, \cdots$ be a sequence of projections on a normed space $X$. Suppose that $P_{n+1} P_n = P_n$ for all $n$ and that the union of the ranges of these projections is dense in $X$. Suppose further that $\sup_n \|P_n \| < \infty$. Prove that $P_n x \rightarrow x, > \forall x \in X.$ [Problem 4.4.3, p.197.]
Thus, $X$ is not even assumed to be a Banach space. These are my thoughts on this:
First, let $V_n$ be the range of each $P_n$. Thus, $X = \overline{ \bigcup _{n=1}^\infty V_n}.$ It is easy to demonstrate that $V_i \subset V_j, j>i:$ $$Let \, \, v_i \in V_i \Rightarrow P_iv_i =v_i.\,\, Then\,\, P_{i+1}(P_i(v_i))=P_iv_i = v_i \Rightarrow v_i \in R(P_{i+1})=V_{i+1}.$$ It is also easy to show that if $x \in V_n \Rightarrow, \exists N,\,s.t.\, P_{N \geq n}x =x.$ Thus, if $x\in \bigcup _{n=1}^\infty V_n \subset X$, then as $n \rightarrow \infty$, $\exists N$ such that $P_nx=x, \forall n \geq N.$
But what if $x \in X \setminus \bigcup _{n=1}^\infty V_n $? I understand that in that case, $x$ would be a limit point of the union of the ranges, due to $ \bigcup _{n=1}^\infty V_n$ being dense in $X$.
I cannot push it much further from here though. I'd appreciate any help or suggestions for better approaches.
Let $x \in X$ and $\epsilon >0$. There exist $y \in \bigcup_n V_n$ such that $\|x-y\| <\epsilon$. Now $\|P_n x-x\| \leq \|P_n x-P_ny\|+\|P_ny-y\|+\|y-x\|\leq (M+1)\epsilon +\|P_n y-y\|$ where $M=\sup_n \|P_n\|$. Can you finish?