The kernel of a Fredholm operator is not continuous with respect to the Hausdorff distance between the spheres: 

For $t\geq 0$, consider the operator $S_t:X\times Y \to X\times Y$ defined by $S(x,y)=(tx,y)$ where $X,Y$ are Banach spaces and $X$ is finite dimensional. 

When the operator $T:X\to Y$ has closed range, you have semicontinuity in some sense. See Proposition 3.1 
[here.][1]

When the operator $T:X\to Y$ has non-closed range and it is injective, you can find a compact perturbation $K:X\to Y$ with arbitrarily small norm so that the kernel of $T+K$ is infinite dimensional. So there is no semicontinuity.   

  [1]: https://www.researchgate.net/publication/243033124_The_gap_between_subspaces_and_perturbation_of_non-semi-Fredholm_operators