Construct a normed space $X$ and a bounded linear operator $T \colon X \rightarrow X$ with $\|T\|<1$ such that the equation $x- T(x) = y$ has no solution for some $y \in X$.
I know that if $B$ is a Banach space and $T \colon B \rightarrow B$ is a bounded linear operator with $\|T\|<1$, then the equation $x - T(x) = y $ has a unique solution $x \in B$ for every $y \in B$ that depends continuously on $y$. But I don't know how to build a normed space in which said equation has no solution. I appreciate any help.
Take $X=\mathcal{P}(x)$ the space of polynomials with uniform norm on $ [0,1].$ Let $$Tp= {x\over 2}p.$$ Then the equation $p-Tp=1$ has no solutions