I'm trying to show the following: if $H_1, H_2$ are separable Hilbert spaces and $T: H_1 \to H_2$ and $S: H_2 \to H_1$ are bounded linear maps, and $ST = I_1 - E_1$, where $I_1$ is the identity on $H_1$ and $E_1$ is a compact operator $H_1 \to H_1$. Then, the image of $T$ is closed.
We want to show that $y_n = T x_n \to y \in H_2$ means that $y = Tx$ for some $x \in H_1$. I can show it if we assume that $x_n$ are bounded: then since $E_1$ is a compact operator, there's a subsequence $x_{n_k}$ such that $E_1 x_{n_k} \to x$ in $H_1$, and thus $y = T(\lim_k x_{n_k}) = T(\lim_k ST x_{n_k} + E_1 x_{n_k}) = T(Sy + x)$.
Is there a way to reduce to this case when $\|x_n\|$ are not uniformly bounded? One hint I have is that apparently for some $c> 0$, $\|Tx\| \geq c \|x\|$ if $x \in \mathrm{ker}(T)^\perp$, but I'm not sure why this is helpful or how to prove it.
You can use that $S y_n = STx_n = x_n - E_1 x_n \rightarrow S y$, i.e. $x_n - E_1 x_n$ is convergent. Now, decompose $E_1 x_n = v_n+w_n$ with $w_n \in \ker(T)^\perp$. This gives $$TSy = \lim_{n \rightarrow \infty} T(x_n-E_1x_n) = y - \lim_{n \rightarrow \infty} Tw_n.$$ The hint implies that $w_n$ is bounded. Next, we have $$E_1 w_n - w_n = ST w_n \rightarrow S(y-TSy)$$ and we can take a subsequence such that $E w_{n_k} \rightarrow z$. Using the second-last convergence shows that $w_{n_k}$ is already convergent with limes $\omega$. Thus, the third-last equation shows that $$TSy=y - Tw$$ and thus $y= T(Sy+w)$.