I need to know if what I do is right. And if what I do is ok, how to obtain "or $f(x)\geqslant c$, for all $x\in S_r$".
This is the exercise in my list that I'm solving:
Let $x_0\in S_r=\{x\in X;\|x\|=r\}$, where $X$ is a real normed space. Show that there exists a hyperplane $H=\{x\in X;f(x)=c\}$, with $f\in X^\prime$, which contains $x_0$ and keep the sphere $S_r$ in one of his sides, i.e., or $f(x)\leqslant c$, for all $x\in S_r$, or $f(x)\geqslant c$, for all $x\in S_r$.
My partial solution: Since $x_0\in S_r$, we have $\|x_0\|=r$, which implies $x_0\ne 0$. Then, it follows by a corollary of Hahn-Banach Theorem that there exists $f\in X^\prime$ such that $$\|f\|=1\qquad\text{and}\qquad f(x_0)=\|x_0\|.$$ Now, let $$H:=\{x\in X;f(x)=r\}.$$ We have that $x_0\in H$, as required, because $f(x_0)=\|x_0\|=r$. Besides, for all $x\in S_r$, it follows that $$f(x)\leqslant|f(x)|\leqslant\|f\|\|x\|=1\cdot r=r$$ $$\Longrightarrow\quad f(x)\leqslant r,\ \forall x\in S_r.$$ And we are almost done. But, what about the "or $f(x)\geqslant r,\forall x\in S_r$"?
You are done. You needed to satisfy $f(x) \le r$ for all $x \in S_r$, or you needed to satisfy $f(x) \ge r$ for all $x \in S_r$. You don't need to satisfy both at the same time.
If you wanted to obtain the other condition, you could arbitrarily have taken $g = -f$, and $s = -r$. Then $$x \in S_r \implies f(x) \le r \implies -f(x) \ge -r \implies g(x) \ge s.$$ So, we have a different functional where the inequality is swapped. The actual hyperplane has not been altered, but you can see why $\le$ or $\ge$ are interchangeable here, so long as you stick with the same one for all $x \in S_r$.