In proof of "The Hyperplane $H=\left [ f =\alpha \right ]$ is closed iff $f$ is continuous" we choose $x_0 \in H^c$ so that $f(x_0) \neq 0$,for example,$f(x_0)<\alpha$ then we prove $f(x) < \alpha$,$\forall x \in B(x,r) \subset H^c$ and $f$ is continuous.
But if $f(x_0) >\alpha$,what happen? If $f(x) > \alpha$,$\forall x \in B(x_o,r)$ then $f$ is not continuous?
See more at:in normed space hyperplane is closed iff functional associated with it is continuous
The "for example" in the linked proof should be read as "without loss of generality": An entirely similar proof shows that if $f(x_{0}) > \alpha$ and $B(x_{0}, r) \subset H^{c}$, then $f(x) > \alpha$ for all $x$ in $B(x_{0}, r)$.
Alternatively, if you prefer, the case $f(x_{0}) > \alpha$ reduces to the case $-f(x_{0}) < -\alpha$, i.e., by replacing the linear functional $f$ with $-f$; obviously $-f$ is continuous if and only if $f$ is continuous.
