If E is a Hilbert space and $T \in B(E)$ is compact, show $T(E)$ does not contain a closed infinite dimensional subspace

Question

It's the problem from "Essential Results of Functional Analysis," R.J. Zimmer, Chapter 3, problem 3.1. I try to prove this problem and I am confused with the condition "closed infinite dimensional." Consider a compact operator T, given by diagonal matrix $T_{ij} = \lambda_i\delta_{ij}$, with $\lambda_i \to 0$. Then for Hilbert space $E$, we have $T(E) = E$. That means $T(E)$ has a closed infinite dimensional subspace($E$ itself). Where did I go wrong?

Answer 1

Hint: For each closed subspace $F \subset E$, there is an orthogonal projection $P_F$ onto $F$. The composition $P_F \circ T$ is also compact. So if $T(E)$ contained a closed infinite-dimensional subspace $F$, we'd have a compact map onto an infinite-dimensional Hilbert space. Why can that not happen?