Hahn-Banach dominated extension theorem for real vector spaces

Let $\mathcal{E}$ be the set of all pairs $(N,g)$ such that $M\subseteq N\subseteq X$, the set $N$ is a linear subspace of $X$, the map $g:N\to \mathbb{R}$ is linear, $g{\mathrm{\mid }}_M=f$, and $g\le p$ on $N$. Order $\mathcal{E}$ by extension: $(N_1,g_1)⪯(N_2,g_2)⟺N_1\subseteq N_2\text{ and }{g}_2{\mathrm{\mid }}_{N_1}=g_1\mathrm{.}$ The pair $(M,f)$ lies in $\mathcal{E}$, so this poset is nonempty.

Let $\mathcal{C}\subseteq \mathcal{E}$ be a chain. If $\mathcal{C}=\mathrm{\varnothing }$, then $(M,f)$ is an upper bound for it. If $\mathcal{C}\mathrm{\ne }\mathrm{\varnothing }$, [L2] applies to the union of its domains and yields a well-defined linear functional dominated by $p$; because every chain element extends $f$, that union functional still extends $f$. Hence every chain in $\mathcal{E}$ has an upper bound in $\mathcal{E}$.

By [L3], choose a maximal element $(N,F)$ of $\mathcal{E}$. If $N\mathrm{\ne }X$, choose $x_0\in X\setminus N$. Applying [L1] to the dominated functional $F$ on the subspace $N$ produces a dominated linear extension $\tilde{F}$ on $N+\mathbb{R}{x}_0$. Then $(N,F)\prec (N+\mathbb{R}{x}_0,\tilde{F})$, contradicting maximality. Therefore $N=X$.

Since the maximal domain is all of $X$, the corresponding functional $F$ is the required dominated extension of $f$.