= Solution
\b[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 $y^{(N)}=L(1,\ldots,1,0,\ldots)$ 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 \b[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>.
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