Let $X$ be a closed proper subspace of $V$. Prove $\forall \alpha\in (0,1), \exists x\in V$ such that $\|x\|=1$ and $\inf_{y\in X}\|Tx-Ty\|\geq\alpha$ Let $(V,||\cdot||)$ be a normed space and $X\subset V$ a closed, proper subspace. Let $T\in B(V)$ such that $(1-T)V\subset X$. Prove that, given $\alpha\in (0,1)$, there exists $x\in V$ with $\|x\|=1$ such that $\inf_{y\in X}\|Tx-Ty\|\geq \alpha$, where
$$B(V)=\{T:V\to V; T \text{ is bounded}\}$$
First thing that came to mind is Riesz's Lemma. Since $X$ is a closed proper subspace of $V$, then for $\alpha\in (0,1)$, there exists $x\in V$ such that $\|x\|=1$ and $\|x-y\|\geq \alpha$ for all $y\in Y$, so I only need to prove $\inf_{y\in X}\|Tx-Ty\|\geq \alpha$. How to use the fact that $T$ being a bounded operator implies $(1-T)V\subset X$?
 A: First I will restate the question with the notation I'm going to use because the notation established in the original post confuses me.

Question: Let $(X,\|\cdot\|)$ be a normed space and $Y\subset X$ a closed, proper subspace. Let $T\in B(X)$ be a bounded operator such that $(1-T)X\subset Y$. Show that, given $\alpha\in(0,1)$, there exists $x\in X$ with $\|x\|=1$ such that $\|Tx-Ty\|\ge\alpha$ for all $y\in Y$.

Proof: First, note that $(1-T)Y\subset (1-T)X\subset Y$, so $y-Ty\in Y$ for all $y\in Y$. Write $Ty=y-(y-Ty)\in Y$ to conclude that $T(Y)\subset Y$. Since $Y$ is closed and proper, we now apply Riesz lemma to obtain $x\in X$ with $\|x\|=1$ such that $\|x-y\|\ge\alpha$ for all $y\in Y$, so $\inf_{y\in Y}\|x-z\|\ge\alpha$.
Recall the quotient norm on $X/Y$: we have $\|x+Y\|=\inf_{y\in Y}\|x-y\|$. Now since $(1-T)X\subset Y$, we have that $x-Tx\in Y$, so $Tx+Y=x+Y$, so $\|Tx+Y\|_{X/Y}=\|x+Y\|_{X/Y}$, so $\inf_{y\in Y}\|Tx-y\|=\inf_{y\in Y}\|x-y\|\ge\alpha$. Therefore we have $\|Tx-y\|\ge\alpha$ for all $y\in Y$. Since $T(Y)\subset Y$ we conclude that $\|Tx-Ty\|\ge\alpha$ for all $y\in Y$, as we wanted.
