The application of Hahn-Banach Theorem 02

$X'$ = $\{ f: X \to \mathbb{K} \}$ where $f$ is is linear and continuous and $X$ is a Banach space over $\mathbb{K}$. Prove that $X' \neq {0}$, in fact, for every $x \neq 0 \in X$, we can find $f \in X’$ such that $f(x) = |x|$ and $|f| = 1$.

Proof: Pick $x_0 \in X$. Define $X_0 = x_0 \cdot \mathbb{K}$, a subspace of $X$, and $g: X_0 \to \mathbb{K}$, $g(x) = x$, which is linear. Since $g$ and $|\cdot|$ satisfy the conditions of the Hahn-Banach theorem, we can find $f: X \to \mathbb{K}$ such that $f|_{X_0} = g$, $f$ is linear and $f(x) \leq |x|$, $\forall x \in X$. Therefore $f(x_0) = g(x_0) = |x_0|$ and $|f| \leq 1$. The equality $f(x_0) = |x_0|$ guarantees that $|f| = 1$.