Recall that a Hilbert $B$-module $X$ over a $C*$-algebra $B$ is completion of a right $B$-module $E$ endowed with a $B$-valued inner product such that $(x, x) \geq 0$, $(x, x) = 0 \iff x = 0$, $(x, y)^* = (y, x)$, and $(x, yb) = (x, y)b$ for all $x, y\in X$ and $b \in B$. The completion is with respect to the norm $\|x\|_X = \|(x, x)\|_B^{1/2}$. Let $T$ be a $B$-linear map from $X$ to itself, with this $B$-valued inner product, one can define the adjoint $T^*$ in the usual sense: $(Tx, y) = (x, T^*y)$ and the norm of $T$ is the operator norm in the usual sense as well. We can define the rank-one operator $x\otimes y: z \mapsto x(y, z)$ for $x, y, z \in B$ and we write $K(X)$ to be the closed span of rank-one operators.
Now my question is that: if the identity operator $1_X$ is compact, then how to show that $1_X$ is a linear combination of finite rank-one operators? I could not find any proof of it but it seems a well-known fact that was used in many papers.
Note that $FR(X)$, the linear span of the rank one operators, is a dense ideal in $K(X)$. Thus, in case $K(X)$ is unital, there is some $T\in FR(X)$, with $$ \|T-1_X\|<1. $$ By a well known result, $T$ must be invertible, but if an ideal contains an invertible element, it must coincide with the whole algebra.