A compact map $c \colon H \to H'$ between Hilbert spaces is closed when restricted to closed and bounded subsets. I'm trying to prove the following claim:

Let $L \colon H\to H'$ be a continuous linear Fredholm map between Hilbert spaces, and let $c \colon H\to H'$ be a compact map. Prove that $L+c \colon H \to H'$ is proper when restricted to any closed bounded subset $A \subset H$.

The idea of the proof given was to consider the following factorisation of the map. First we consider $$h\colon A \to H' \times \overline{c(A)}\times \overline{\rho(A)}$$ $$a\mapsto (L(a),c(a),\rho(a))$$ where $\rho(A)$ is the orthogonal projection $H\to ker(L)$. Then compose with the homeomorphism 
$$ \phi \colon   H' \times \overline{c(A)}\times \overline{\rho(A)} \to H' \times \overline{c(A)}\times \overline{\rho(A)}$$ $$ (a,b,c)\mapsto (a+b,b,c)$$ and then projecting on the first coordinate. The last two maps are easily seen to be proper, so the only thing that remains to be proved is properness of the first one.

The claim is that this map is proper since injective and CLOSED. Injectiviness is clear, I've problems proving that this map is closed, in particular that the map $$ c \colon A \to \overline{c(A)}$$ $$a \mapsto c(a)$$ is closed since the other two components are closed.

Can someone help me see why is that the case?
 A: I'm not an expert in functional analysis I fear what you want to prove is somehow "wrong" You have to use the fact that $L$ is Fredholm. Nonetheless the claim about about $$h\colon A \to H' \times \overline{c(A)}\times \overline{\rho(A)}$$ $$a\mapsto (L(a),c(a),\rho(a)$$ being a closed map is true. Since any closed subspaces of $A$ are gonna be closed and bounded, we might as well work with $A$ directly.
Let $\{(L_n,c_n,\rho_n)\}_n\subset \text{Im}(h)$, and assume that $$(L_n,c_n,\rho_n)\to (L_{\infty},c_{\infty},\rho_{\infty}) \in \overline{\text{Im}(h)}$$ Clearly if we prove that $(L_{\infty},c_{\infty},\rho_{\infty}) \in \text{Im}(h)$ we are done. Let $\{a_n\}_n \in A$ such that $h(a_n)= (L_n,c_n,\rho_n)$.
Recall the following lemma about Fredholm operators:
Lemma: Let $L\colon X \to Y$ be a linear operator between Banach spaces. TFAE
1) $\ker(L)$ is finite dimensional and $\text{Ran}(L)$ is closed
2) Every bounded sequence $\{x_i\}$ in the domain of $L$ with $\{L(x_i)\}$ convergent has a convergent subsequence.
This is lemma 16.17 here. Since $L$ is Fredholm in our case, point (1) applies, hence by the Lemma point (2) is true. Since $L(a_n)\to L_{\infty}$, by (2)  (the sequence is of course bounded) we can extract a  convergent subsequence $a_n'\to a$ and since $A$ is closed, $a\in A$. Now it's immediate to see that by the uniqueness of the limit, $$h(a)=(L_{\infty},c_{\infty},\rho_{\infty})$$ which conclude the proof since we have $$\overline{\text{Im}(h)}=\text{Im}(h)$$
