Initially I want to clarify that the idea is to prove this theorem without using Zorn's Lemma. (and therefore without using the axiom of choice, since these are equivalent.)
I know this is a somewhat hackneyed question. Still, I want to present the demonstration that I have written, about which I feel there are some details to be clarified, and that is where I need the go-ahead from a second observer.
The proof of the Hanh-Banach theorem is proposed in Kreyszig's book - Introductory Functional Analysis with Applications, more precisely, it is exercise number 9 of section 4.3, which is on page 224 of this text. It should be noted that it is possible to use results previously demonstrated in this section of the book for the demonstration of the exercise in question.
I want to highlight that some details regarding the notation and/or nomenclature in the following sketch are subject to theorem 4.2-1 of the book in question (Hahn-Banach theorem. Extension of linear functionals), which is in the page 214. I also attach the following link where you can view the text:
Hanh-Banach theorem (separable normed spaces.) Let $f$ be a bounded linear functional defined in a subspace $Z$ of a separable normed space $X$. Then there exists a bounded linear functional $\widetilde{f}$ in $X$, which is an extension from $f$ to $X$ and has the same norm, $$\|\widetilde{f}\|_X=\displaystyle{\sup_{{x\in X\atop\|x\|=1}}} |\widetilde{f}(x)|, \hspace {.2cm}\|f\|_Z=\displaystyle{\sup_{{x\in X\atop\|x\|= 1}}} |f(x)|. $$ (and $ \|f\|_Z=0$ in the trivial case $ Z=\{0\}).$
Sketch of the proof: Let $y_1\in X \setminus Z$. For each $c\in\mathbb{R}$ we can obtain a linear extension of $f$ to the space $Z_1$, generated by $Z$ and $\{y_1\}$ by means of $g_1(z + ay_1)=f(z)+ac$ and we can choose $c$ such that $g_1(x)\leq p(x) $ for all $ x\in Z_1$.
If $X\neq Z_1$, we can take $y_2\in X\setminus Z_1$ and repeat the process, and thus obtain an extension of $f$ to the subspace $Z_2$ generated by $Z_1$ and $\{y_2\}$, etc. Obtaining in this way a sequence of subspaces $Z_j$ such that $Z_1\subset Z_2\subset\cdots$ and such that $f$ can be extended from one to the next and the extension $g_j$ satisfies $g_j(x)\leq p(x)$ for all $x\in Z_j$.
If $X=\bigcup_{j=1}^n Z_j $ we reach the desired extension $\widetilde{f}$ in $n$ steps.
If $ X=\bigcup_{j=1}^\infty Z_j $ we can use ordinary induction and get the desired result.
