A confusion about an exercise about an affine subspace that contains a hyperplane I am solving this problem from Brezis's book of Functional Analysis.

Let $E$ be a normed vector spacr and $H$ a hyperplane, i.e., $H=\{x\in E: f(x)= \alpha\}$ for some linear functional $f$ and some real number $\alpha$. Let $V$ be an affine subspace (i.e., $V=U+a$ where $a \in E$ and $U$ is a vector subspace of $E$) that contains $H$.  Prove that either $V=H$ or $V=E$.

I found a proof here but could not understand it. On the other hand, It seems I have found a counter example.

Let $E$ be infinite dimensional, $f$ injective continuous, and  $\alpha=0$. This means $H = \{0\}$. Fix $0 \neq y \in E$ and let $V := \operatorname{span} \{0, y\}$. Then $H \subsetneq V \subsetneq E$.

Could you confirm if I understand the exercise properly?
 A: Note that this a purely algebraic result, the norm is irrelevant here.
Also note that a functional cannot be injective in a space with dimension $>1$. Suppose $u,v$ are linearly independent, then both $f(u), f(v)$ are non zero and so $f( {1 \over f(u)} u ) = f( {1 \over f(v)} v ) = 1$ which would contradict injectivity.
If $f=0$ then either $H=E$ if $\alpha=0$ and we are done, or
$H= \emptyset$ if $\alpha \neq 0$ in which case the result is false.
So suppose $f \neq 0$.
If $V \neq H$ then there is some $v \in V \setminus H$. We must have $f(v) \neq \alpha$.
Pick $x \in E$. If $f(x) =\alpha$, then $x \in V$, so suppose $f(x) \neq \alpha$.
If $f(x) \neq f(v)$, then there is some $t \neq 0$ such that $f(tx+(1-t)v) = \alpha$ and so $tx+(1-t)v = h$ for some $h \in H$. Then
$x = {1 \over t} h + (1-{t \over t}) v \in V$.
Finally, if $f(x) = f(v)$, then if $h_1 \in H$ then $f(h_1 +x-v) = \alpha$
and so $h_2 = h_1 +x-v$ and $x = h_2 -h_1+v$ and so $x \in V$.
Hence $V=E$.
A: First, $V = H$ is equivalent to $U = H':= \{x \in E \mid f(x) = \alpha- f(a)\}$. Second, $V = E$ is equivalent to $U = E$. Third, $V$ contains $H$ is equivalent to $U$ contains $H'$. Then our problem reduces to

Let $E$ be a normed vector space, $f$ a linear functional, and $\alpha \in \mathbb R$. Let $V$ be a vector subspace that contains $H:=\{x \in E \mid f(x) = \alpha\}$. Prove that either $V = H$ or $V=E$.

Assume the contrary that $H \subsetneq V \subsetneq E$. Then there are $x_0 \in V \setminus H$ and $x_1 \in E \setminus V$. If $f(x_1) =0$ for all $x_1 \in E \setminus V$, then $H + E \setminus V \subseteq H \subsetneq V$ which is a contradiction. WLOG, we assume $f(x_1) \neq 0$. Notice that $f (x_0) \neq \alpha$. We find $\lambda \neq 0$ such that $f (x_0 + \lambda  x_1) = \alpha$. This is possible with $\lambda = (\alpha-f(x_0)) / f(x_1)$. It follows that $x_0 + \lambda  x_1 \in H \subsetneq V$. On the other hand, $x_0 \in V$. This implies $x_1 \in V$ which is a contradiction. This completes the proof.
