Let $H$ be a Hilbert space and $P$ be a projection to a finite dimensional subspace $K$ of $H$, for a $T\in B(H)$, if $||PTP||=1$, then, for arbitrary $\epsilon>0$, there exists a vector $\alpha$ such that $||T \alpha-\alpha||<\epsilon$.
How to prove it? Or it is a wrong proposition.
If you do not need any control over the norm of the vector $\alpha$, then, yes, such a vector exists. Take any vector with norm less than $\min (\frac{\varepsilon}{2}, \frac{ \varepsilon}{ 2 \Vert T \Vert})$.
If you add the assumption that $\Vert \alpha \Vert=1$, such a vector $\alpha$ need not exist. Consider the operator $T(x)=-x$. Its restriction to any finite dimensional subspace has norm 1 and $\Vert Tx -x \Vert = 2\Vert x \Vert$ for any $x \in H$.