The answer involves two arguments (I guess that $Fred (\ell^2)$ does have the norm topology). 

1. If $F: X \to Fred(\ell^2)$ is continuous, then there is another map $G: X \to Fred(\ell^2)$ such that $FG-1$ and $GF-1$ are maps into $\mathcal{K} (\ell^2)$. 

2. If $T:X \to \mathcal{K}(\ell^2)$, then $T$, viewed as an $C(X)$-linear operator on $H_{C(X)}$ is $C(X)$-compact. 

Ad 1: Let $\mathcal{Q}(\ell^2) = \mathcal{B}(\ell^2)/\mathcal{K}(\ell^2)$ be the Calkin algebra and let $\pi: \mathcal{B}(\ell^2) \to \mathcal{Q}(\ell^2)$ be the quotient map. It is well-known that $F \in \mathcal{B}(\ell^2)$ is Fredholm iff $\pi(F)$ is invertible, and also that $\mathcal{Q}$ is a Banach algebra. Let $\mathcal{G} \subset \mathcal{Q}(\ellHence the inversion map