The kernel of a Fredholm operator is not continuous with respect to small norm perturbations:

  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.

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.

The kernel of a Fredholm operator is not continuous with respect to small norm perturbations: 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.

However, when the operator $T:X\to Y$ has closed range, you have semicontinuity in some sense. See Proposition 3.1 here. And from this result (and finite dimension for the kernels) you can get a positive answer to your question.

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.

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.

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.

The kernel of a Fredholm operator is not continuous with respect to small norm perturbations:

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.

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.