Compact operators between Hilbert spaces

Question

I have the suspect that the following statement is true, but I don't how to prove it. Any suggestion? Thanks to all!

Let $X$, $Y$ be Hilbert spaces and let $T \colon X \to Y$ be a linear continuous injective map. Suppose that for every $\epsilon > 0$ there exists a closed vector subspace $V_\epsilon \subseteq X$ of finite codimension such that $\Vert Tv \Vert_Y \leq \epsilon \Vert v \Vert_X$ for all $v \in V_\epsilon$. Then $T$ is compact.

Answer 1

Let $P_\epsilon$ be the orthogonal projection onto $V_\epsilon$ and $$ T_n := T(\mathbb I - P_{1/n}) $$ $T_n$ is a finite-rank operator and $$ \lVert (T - T_n)v \rVert = \lVert T P_{1/n} v \rVert \leq \frac 1 n \lVert v \rVert $$ and so $$ \lVert T - T_n \rVert \leq \frac 1 n \to 0 $$ We can conclude $T$ is compact because it is limit in the operator norm of a sequence of finite-rank operators.