Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.
If $M$ is bounded, it is not difficult to show that $M^2$ is bounded. What about the converse?
Aiming to apply Closed graph, I tried assuming that $x_n \to x$ and $Mx_n \to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$, $$ M^2 x_n \to M^2x, \quad M^3x_n \to M^2y, $$ which does not seem very fruitful. I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.
I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.
Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?
Could one argue as follows:
Let $\{x_i\}_{i\in I}$ be a normalised Hamel basis for $X$. Let $\{y_n\}_{n=1}^{\infty}$ and $\{z_n\}_{n=1}^{\infty}$ be two countable subsets of the basis which are disjoint and such that $y_n\neq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:
$$My_n = nz_n,\quad Mz_n = 0$$
and $Mx_i = 0$ if $x_i\neq z_n$ or $x_i\neq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.