= Solution
On the one-dimensional <linear subspace> $Y=\mathbb R x_0$, define $g(tx_0)=t\|x_0\|$. It is linear and obeys $g(y)\le\|y\|$ 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 $f(x)\le\|x\|$ for every $x$.
Apply this also to $-x$ to get $-f(x)\le\|x\|$. Hence $|f(x)|\le\|x\|$, proving <continuity> and $\|f\|\le1$. At $x_0\ne0$ the extension satisfies $f(x_0)=\|x_0\|$, forcing the reverse <norm> inequality. Therefore
$$
\boxed{f(x_0)=\|x_0\|,\qquad\|f\|=1}.
$$
The exclusion of $x_0=0$ is what ensures that this particular construction attains <norm> one.
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