Proof. Since $F$ is closed and $E$ is locally convex, there exists a continuous seminorm $\rho: E \to [0, \infty)$ such that $\rho(F) = \bracsn{0}$ and $\rho(y) \ne 0$. By the Hahn-Banach Theorem, there exists $\phi \in E^{*}$ such that $F \subset \ker(\phi)$ and $\dpn{y, \phi}{E}= 1$. In which case, $\phi$ induces an isomorphism

whose restriction to $F + Ky$ is an isomorphism onto $F \times K$.

Let $\angles{x_\alpha + \lambda_\alpha y}_{\alpha \in A}\subset F + Ky$ be net convergent in $E$. For each $\alpha \in A$, $\dpn{x_\alpha + \lambda_\alpha y, \phi}{E}= \lambda_{\alpha}$. By continuity of $\phi$, $\angles{\lambda_\alpha}_{\alpha \in A}$ is convergent, and $\angles{x_{\alpha}}_{\alpha \in A}$ is also convergent. Since $\angles{x_\alpha}_{\alpha \in A}\subset F$ and $F$ is closed, $\lim_{\alpha \in A}x_{\alpha} \in F$. Therefore $\lim_{\alpha \in A}(x_{\alpha} + \lambda_{\alpha} y) \in F + Ky$, and $F + Ky$ is also a closed subspace.$\square$