From Brezis Functional Analysis pg.5
Proposition: Let $f$ be a linear functional on $E$ which does not vanish identically, $\alpha \in \mathbf{R}$. Then the affine hyperplane $H:=\{x\in E\,|\,f(x)=\alpha\}$ is closed if and only if $f$ is continuous.
My Proof: If $f$ is continuous and $\{h_k\}$ a sequence in $H$ then $\alpha = f(h_k)\rightarrow f(h) = \alpha$ as $k\rightarrow \infty$ and $H$ is closed.
Conversely let $H$ be closed and let $\{h_k\}$ be a sequence converging to $h\in H$. Then $f(h)=\alpha$ and $$ \lim_{k\rightarrow \infty}f(h_k) = \lim_{k\rightarrow \infty}\alpha = \alpha = f(h) $$ so $f$ is continuous. $\square$
Brezis does something far different for the second part and I was wondering if I have made a very fundamental mistake or he has chosen his method for pedagogical reasons. Thanks! Also any thoughts on my presentation and proof writing are appreciated.
Resolution: Guido has found the error. I have only proved that $f$ is continuous on $H$ where I actually need to prove that $f$ is continuous on all of $E$.
You have shown that $f$ is continuous in $H$, but that will happen regardless of $H$ being closed simply because $f|_H$ is constant. What you have to prove is that $H$ being closed implies continuity of $f$ in its domain (and not only restricting it to $H$).
Here's another proof:
Note that translation by a point is a homeomorphism, so $H - p$ will be closed for any $p \in E$. Now, if $f$ is zero the result follows, and otherwise by dimension $f$ is surjective and thus we can choose $x \in f^{-1}(\alpha)$. It follows that $\ker f = H - x$ is closed. Therefore, we can assume $\alpha = 0$ and $H = \ker f$.
Taking $y \in E \setminus \ker f$ , if $f$ were not bounded then we should have a sequence of unit vectors $x_n$ such that $f(x_n) \geq n$ for each $n \in \mathbb{N}$. But then it is
$$ f(y)f(x_n)^{-1}x_n -y \to y $$
with $f(y)f(x_n)^{-1}x_n -y \in \ker f$, and that contradicts the fact that $\ker f$ is closed.