$T$ is a compact operator on a Hilbert space. Show that $\operatorname{im}(T)$ does not contain a closed infinite dimensional subspace.
Here is my attempt at the problem: Suppose that $\operatorname{im}(T)$ had a closed infinite dimensional subspace. Then let $\{e_n\}$ be its orthonormal basis. since $||e_n-e_m||=2$ for all $n,m$, it follows that the infinite sequence ${e_n}$ does not have a convergent subsequence. which would mean that a closed subset of a compact set is not compact.
Assume that $im(T)$ is closed and infinite-dimensional. Consider the quotient space $\hat T=X/\ker T$. Then the operator $$ \hat T:\hat X\to im(T), \ \hat T(\hat x)= Tx, \ x\in \hat x $$ is bounded and continuous. Moreover, since $\ker T$ is closed, $\hat X$ is a Banach space. Then $\hat T$ is a linear, continuous, and bijective mapping, thus it has a continuous inverse. However, since $T$ is compact, $\hat T$ is compact as well. Contradiction.