Every nonzero vector has a norming functional

Let $M:=\mathrm{span}\{x\}$. Define $f_0:M\to \mathbb{R}$ or $f_0:M\to \mathbb{C}$ by $f_0(\lambda x):=\lambda \mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$ Because $x\mathrm{\ne }0$, each vector of $M$ has a unique representation $\lambda x$. Moreover, $\mathrm{\mid }{f}_0(\lambda x)\mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }\lambda x\mathrm{\parallel },$ so $f_0$ is bounded and $\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$.

If the scalar field is $\mathbb{R}$, [L2] extends $f_0$ to a bounded real linear functional $f$ on $X$ with $\mathrm{\parallel }f\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$. Since $x=1\cdot x$, $f(x)=f_0(x)=\mathrm{\parallel }x\mathrm{\parallel }$.

If the scalar field is $\mathbb{C}$, [L3] extends $f_0$ to a bounded complex linear functional $f$ on $X$ with $\mathrm{\parallel }f\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$. Again $f(x)=f_0(x)=\mathrm{\parallel }x\mathrm{\parallel }$.

In either scalar case, the extension produced in step 2.1 or step 2.2 is a bounded linear functional on $X$, hence an element of $X^{\ast }$ by [L1], with norm $1$ and value $\mathrm{\parallel }x\mathrm{\parallel }$ at $x$.