Let $X$ be a complex Banach space and $T$ be a bounded linear operator on $X$. Put $Y=\{x\in X:\|Tx\|^2\leq\|T^2x\|\|x\|\}$. Is $Y$ a subspace of $X$?
I know is that $Y$ is closed and $aY$ is contained in $Y$ for all scalar $a$. The only thing which needs explanation is that if $x$ and $y$ are in $Y$, then $x+y$ is in $Y$. In all the examples which I worked out Y turned out to be a subspace. Therefore, my guess is that the answer should be affirmative.
You look in the wrong direction. Take $X = \mathbb{C}^5$, and look at $T$ given by the matrix
$$\begin{pmatrix}0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & -1 & 0 \end{pmatrix}.$$
Then $e_1, e_2 \in Y$, but $e_1 + e_2 \notin Y$, since
$$\lVert T(e_1+e_2)\rVert^2 = \lVert e_3 + e_4\rVert^2 = 2 > 0 = \lVert T^2(e_1+e_2)\rVert\cdot \lVert e_1+e_2\rVert.$$