If $X$ is a complex Banach space, is the set $T \in L(X)$ with finite dimensional kernel dense? If $X$ is a complex Banach space, is the set $T \in L(X)$ with finite dimensional kernel  dense? Here $L(X)$ is the set of bounded linear operators from $X$ to itself equipped with the norm topology. 
Edit: I am only interested in the separable case. 
 A: No. Not in general.
Consider the space of complex bounded continuous functions on the closed unit ball, denoted $X$. Equip this with the usual topology. It is a separable Banach space. Define $T\in L(X)$ by $(Tf)(x) = f(0)$. We observe that $||T(f)-T(g)|| = |f(0)-g(0)| < ||f-g||$ so the operator is continuous. 
Now, we argue that both $\ker{T}$ and $X\sim\ker{T}$ are infinite dimensional. Take the closed unit ball in $\ker{T}$ and call it $B$. We observe that it is not compact and therefore $\ker{T}$ is not finite dimensional: by Arzela-Ascoli, it suffices to observe that the functions in $\ker{T}$ are not equicontinous as the sequence $x^{n}$ lives in $\ker{T}$.  A modification of the same argument works for $X\sim\ker{T}$.
Let a $U\in L(X)$ with finite dimensional kernel be given. We can take $x$ in the kernel of $T$ and not in the kernel of $U$ by the result of the last paragraph. Set $x^{\prime} = x/||U(x)||$.
Observe that since $x^{\prime}$ is in the kernel of $T$,  $||T-U|| \geq ||T(x^{\prime})-U(x^{\prime})|| = ||U(x)||/||U(x)|| = 1$. This completes the proof.
