 Solution

No. Reuse the diagonal operator on sequence space $L(x_n)=(x_n/n)$ on ${\ell }^2$. Its kernel is zero, hence finite-dimensional. Let $y^{(N)}$ be the truncation of $(1/n)$ to its first $N$ coordinates. Each $y^{(N)}=L(1,\dots ,1,0,\dots )$ lies in the range, while $y^{(N)}\to (1/n)$ in ${\ell }^2$. The limit is not in the range because its formal preimage is not square summable. Thus finite-dimensional kernel does not imply closed range. Equivalently the unit vectors $e_n\in (\ker L{)}^{\perp }$ violate every positive closed-range bound on the kernel complement.