Step 1: One-Dimensional Extension

First, we show how to extend $f$ from $M$ to a subspace $M_1 = M + \text{span}\{x_1\}$ where $x_1 \in X \setminus M$.

For any $\alpha \in \mathbb{R}$, define $F : M_1 \to \mathbb{R}$ by $F(x + \lambda x_1) = f(x) + \lambda \alpha$ for $x \in M$, $\lambda \in \mathbb{R}$. We need $F(x + \lambda x_1) \leq p(x + \lambda x_1)$ for all $x \in M$, $\lambda \in \mathbb{R}$.

For $\lambda > 0$: $f(x) + \lambda \alpha \leq p(x + \lambda x_1)$ implies $\alpha \leq \frac{p(x + \lambda x_1) - f(x)}{\lambda}$

For $\lambda < 0$: Setting $y = -x/\lambda$ and $\mu = -1/\lambda > 0$, we get $\alpha \geq \frac{f(y) - p(y - \mu x_1)}{\mu}$

So we need: $\sup_{y \in M} \frac{f(y) - p(y - x_1)}{1} \leq \alpha \leq \inf_{x \in M} \frac{p(x + x_1) - f(x)}{1}$

Claim: Such an $\alpha$ exists. For any $x, y \in M$: $f(y) - f(x) = f(y - x) \leq p(y - x) = p((y - x_1) + (x_1 + x) - x)$ $\leq p(y - x_1) + p(x + x_1)$

Therefore $f(y) - p(y - x_1) \leq f(x) + p(x + x_1)$, which gives $\sup_{y \in M} [f(y) - p(y - x_1)] \leq \inf_{x \in M} [p(x + x_1) - f(x)]$

Choose any $\alpha$ between these bounds.

Step 2: Zorn's Lemma

Let $\mathcal{F}$ be the set of all pairs $(M', f')$ where $M' \supset M$ is a subspace and $f' : M' \to \mathbb{R}$ is a linear extension of $f$ satisfying $f'(x) \leq p(x)$ for all $x \in M'$.

Order $\mathcal{F}$ by $(M_1, f_1) \leq (M_2, f_2)$ if $M_1 \subset M_2$ and $f_2|_{M_1} = f_1$.

Any chain in $\mathcal{F}$ has an upper bound (take the union of the subspaces and the common extension of the functionals). By Zorn's Lemma, $\mathcal{F}$ has a maximal element $(M_{\max}, f_{\max})$.

Step 3: Maximality Implies Full Extension

If $M_{\max} \neq X$, choose $x_1 \in X \setminus M_{\max}$. By Step 1, we can extend $f_{\max}$ to $M_{\max} + \text{span}\{x_1\}$, contradicting maximality. Therefore $M_{\max} = X$ and $F = f_{\max}$ is the desired extension.