Below is exercise 7 from chaper IV Banach spaces in Lang's Real and Functional Analysis:
Let $F$ be a closed subspace of a normed vector space $E$, and let $v\in E, v\notin F$. Show that $F+ \Bbb{R}v$ is closed. If $E=F+ \Bbb{R}v$, show that $E$ is the direct sum of $F$ and $\Bbb Rv$ (meaning the map $\phi(f,rv)= f+rv$ is a toplinear isomorphism from $F\times \Bbb Rv$ to $E$, i.e. a homeomorphism and isomorphism).
I can prove $F+ \Bbb{R}v$ is closed by looking at the quotient space $E/F$. As the image of $F+ \Bbb{R}v$ under the quotient map $\rho$ is homeomorphic to $\Bbb R$, it is automatically closed in $E/F$, whose inverse image is closed in $E$ by continuity of $\rho$. But $\rho^{-1}(\rho(F+ \Bbb{R}v))=F+ \Bbb{R}v$, thereby proving the closeness of $F+ \Bbb{R}v$. But I am stuck at showing the latter statement. It suffices to show that $\phi$ is an open map, which amounts to showing $U_1+U_2$ is open if $U_1$ and $U_2$ are open subsets of $F$ and $\Bbb Rv$, respectively. Lang mentions that this is an easy consequence of the open mapping theorem, which is a more general result. However, doesn't that assume completeness of $E$? I try to use the quotient space technique, but that doesn't seem to apply here as $U_1+U_2$ needs not be saturated. How should I proceed? Thanks in advance.
Let $\phi:F\times\mathbb{R}v\to E$ be defined by $\phi(f,rv):=f+rv$.
It is continuous since it is the composition of addition and scalar multiplication. It is clearly linear. It is onto by hypothesis, and one-one since $v\notin F$: $$f_1+r_1v=f_2+r_2v\implies f_1-f_2=(r_2-r_1)v$$
Hence $\phi$ is invertible and what remains to be shown is $f+rv\mapsto(f,rv)$ is continuous.
By the Hahn-Banach theorem, since $F$ is closed, there is a continuous functional $\psi$ of unit norm such that $\psi F=0$ but $\psi(v)=t\ne0$. Let $\pi(f+rv):=\psi(f+rv)v/t=rv$. Then $\pi$ is a continuous projection with image $\mathbb{R}v$ and kernel $F$, that is \begin{align*}\|rv\|&=\|\pi(f+rv)\|\le c\|f+rv\|\qquad(c=\|\psi\|\|v\|/t)\\ \|f\|&\le\|f+rv\|+\|rv\|\le(1+c)\|f+rv\|\end{align*} It follows that $E=F\oplus\mathbb{R}v$.