For $V=\{x\in X\mid f(x)=1\}$ show that $\inf\{\Vert x\Vert\mid x\in V\}=\frac1{\Vert f\Vert}$, where $X$ is banach and $f$ is a nontrivial element of the dual space of $X$.
For $x\in V$ we have $$\vert f(x)\vert=1\leq\Vert f\Vert\Vert x\Vert\implies\Vert x\Vert\geq\frac1{\Vert f\Vert}.$$ So in particular $\inf\{\Vert x\Vert\mid x\in V\}\geq\frac1{\Vert f\Vert}$,.
However, I'm not able to show: $$\inf\{\Vert x\Vert\mid x\in V\}\leq\frac1{\Vert f\Vert}.$$
This follows from $$ \|f\| = \sup_{\|x\| = 1} |f(x)|. $$ If you like, take a sequence $x_k$ such that $\|x_k\|=1$ and $|f(x_k)| \to \|f\|$. Then set $y_k = x_k / |f(x_k)|$, so $|f(y_k)| = 1$, and $\|y_k\| \to \frac{1}{\|f\|}$.