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}(\ell^2) $ be the group of units. Hence the inversion map $\iota:\mathcal{G} \to \mathcal{G}$ is continuous. 
By the Bartle-Graves theorem, there is a continuous section $\sigma: \mathcal{Q} \to \mathcal{B}(\ell^2)$. With these notations, define
$$G:= \sigma \circ \iota \circ \pi \circ F: X \to Fred (\ell^2) \subset \mathcal{B}(\ell^2).$$

Ad 2: Let $T:X \to \mathcal{K}(\ell^2)$ be continuous. Let $\epsilon>0$. For each $x \in X$, pick a finite rank operator $R_x$ so that $\| R_x-T_x\| \leq \epsilon /3$, and pick a neighborhood $U_x$ such that $\| T_x-T_y\| \leq \epsilon/3$ for $y \in U_x$. Take a finite subcover $U_1 , \ldots, U_n$ of $(U_x)_{x \in X}$, and a partition of unity $\lambda_1, \ldots, \lambda_n$. Then 
$$\| T_x- \sum_{j} \lambda_j (x) R_{x_j} \|  \leq \epsilon.$$
The operator $\sum_{j} \lambda_j (x) R_{x_j}$ is of finite rank (in the Hilbert module sense), and it follows that $T$ is compact (in the Hilbert module sense).