Let $I$ be the identity operator in $l^2(\mathbb{Z})$ and let $B$ be a compact operator in $l^2(\mathbb{Z})$ and suppose $I+B$ is invertible. $B$ can be written as a bi-infinite matrix with elements $B_{i,j}$ for $i,j \in \mathbb{Z}$: $$ \begin{bmatrix} \ddots & \vdots & \vdots & \vdots & \\ \dots & B_{-1,-1} & B_{0,-1} & B_{1,-1} & \dots \\ \dots & B_{-1,0} & B_{0,0} & B_{1,0} & \dots \\ \dots & B_{-1,1} & B_{0,1} & B_{1,1} & \dots \\ & \vdots & \vdots & \vdots & \ddots\\ \end{bmatrix} $$
Define $B^N$ as a truncated version of $B$ such that its elements are the same as $B$ when both $i$ and $j$ are less than $N$, and equal to zero when at least one of $i$ or $j$ are greater than or equal to $N$.
Is $I+B^N$ invertible?
As $N \to \infty$, we have that $B^N \to B$ so it seems that by taking $N$ large enough we can make $B^N$ as close to $B$ as we want so it 'should' become invertible at some stage. But even if this is the case, this explanation is a bit hand-wavey, how could we state it more rigorously?
Here is a counterexample: Define the entries of $B$ by $$ b_{ij} = \begin{cases} 1/i & \text{ if } j=i, \ i\ge 1,\\ -1/i & \text{ if } j=i-1, \ i \ge 1,\\ 0 & \text{otherwise} \end{cases} $$ So $B$ is equal to $D(I-R)$, where $D$ is diagonal and compact, $R$ is right-shift. So $B$ is compact.
Then the non-trivial rows/columns of $(I+B^n)x=0$ form a linear system of $n$ equations in the unknowns $x_0 \dots x_n$, which has non-trivial solutions. So $I+B^N$ is not injective for all $N$.
Let me argue that $I+B$ is injective (and hence invertible by compactness of $B$): Assume $(I+B)x=0$. Then $x_i=0$ for all $i\le0$, and $$ -i^{-1} x_{i-1} + (1+i^{-1})x_i=0 $$ for all $i\ge1$. Since $x_0=0$ this implies $x_1=0$, and by induction $x=0$.
The claim is true if $\|B^n - B \|_{\mathcal L(l^2)}\to0$. Assume not, then every for $n$ there is $x_n$ with $\|x_n\|=1$ and $(I+B^n)x_n=0$. By reflexivity of $l^2$, $x_n \rightharpoonup x$. Then $$ (I+B)x = (I+B_n-B_n+B)x = (I+B_n)(x-x_n) + (I+B_n)x_n + (B-B_n)x. $$ The right-hand side converges to zero. Hence $x=0$ by injectivity of $I+B$. In addition, $x_n = -B_nx_n \rightharpoonup -x$ and $$ \|B_nx_n-Bx\|\le \|(B_n-B)x_n\| + \|B(x_n-x)\| \to 0. $$ Hence the convergence $-x_n =B_nx_n \to Bx $ is strong, which implies $\|Bx\|=1$, a contradiction to $x=0$.