Let $H$ and $G$ be Hilbert spaces and $T\in B(H.G)$ be a compact operator. Show that if $T$ is surjective, then $\dim(G) \lt \infty$.

Question

I would like to know if my solution is correct or is there an error? And if you know any other way to solve that exercise?

Since $T$ is surjective, $T(H)=G$. Let $W:=\{y\in G|\|y\|\leq 1 \}$ be the closed unit ball of $G$. We will show that $W$ is compact. Let $(y_{n})$ be a sequence in $W$, then there exists a sequence $(x_{n})$ in $H$ such that $(y_{n})=(Tx_{n})$ and $\|y\|\leq 1$. Since $T$ is compact, we have that $(y_{n})=(Tx_{n})$ has a convergent subsequence. Thus $W$ is compact. Hence $\dim(G)<\infty$.