 Solution

On the one-dimensional linear subspace $Y=\mathbb{R}{x}_0$, define $g(t{x}_0)=t\Vert x_0\Vert$. It is linear and obeys $g(y)\le \Vert y\Vert$ even for negative $t$. The norm on $X$ is a sublinear functional, so the real Hahn-Banach theorem extends $g$ to a linear functional $f$ with $f(x)\le \Vert x\Vert$ for every $x$.

Apply this also to $-x$ to get $-f(x)\le \Vert x\Vert$. Hence $|f(x)|\le \Vert x\Vert$, proving continuity and $\Vert f\Vert \le 1$. At $x_0\ne 0$ the extension satisfies $f(x_0)=\Vert x_0\Vert$, forcing the reverse norm inequality. ThereforeThe exclusion of $x_0=0$ is what ensures that this particular construction attains norm one.