= Solution
We prove the required <Hahn-Banach theorem> for a <seminorm> directly. First suppose the <vector space> is real. At an intermediate extension stage, let $g:E\to\mathbb R$ be a <linear functional> with $|g(y)|\leq p(y)$, and let $z\notin E$. Extending to $E+\mathbb Rz$ amounts to choosing $c=f(z)$. The upper domination $f\leq p$ requires
$$
\sup_{y\in E}\{g(y)-p(y-z)\}\ \leq c\leq\ \inf_{w\in E}\{p(w+z)-g(w)\}.
$$
The interval is nonempty: for all $y,w\in E$,
$$
g(y)+g(w)=g(y+w)\leq p(y+w)\leq p(y-z)+p(w+z).
$$
Its endpoints are finite since inserting zero gives bounds $-p(z)$ and $p(z)$, while every lower candidate is below every upper candidate. Choose $c$ in the interval and put $f(y+tz)=g(y)+tc$. For $t>0$, the upper bound with $w=y/t$ proves $f(y+tz)\leq p(y+tz)$. For $t<0$, writing $s=-t>0$ and applying the lower bound with $y/s$ proves the same inequality. The case $t=0$ is already known. Applying this domination to $-x$ and using $p(-x)=p(x)$ gives $|f(x)|\leq p(x)$.
Partially order all dominated <linear functional> extensions of the original $g$ by extension of their domains and values. They form a nonempty set. The union along any chain is a well-defined dominated <linear functional> on a <vector subspace>, so every chain has an upper bound. The <Zorn lemma> gives a maximal extension. If its domain were not $X$, the preceding one-dimensional construction would enlarge it, a contradiction. Thus the required extension exists on $X$.
For a complex <vector space>, apply the proved real result to $\operatorname{Re}g$ on the underlying real <vector subspace>. Let $U:X\to\mathbb R$ be the resulting real <linear functional>, with $|U(x)|\leq p(x)$, and set
$$
f(x)=U(x)-iU(ix).
$$
It is additive and real-linear, and $f(ix)=if(x)$, so it is complex-linear. On $Y$, the identity $\operatorname{Re}g(iy)=-\operatorname{Im}g(y)$ proves $f(y)=g(y)$. Choose $\alpha$ with $|\alpha|=1$ and $\alpha f(x)=|f(x)|$ when $f(x)\ne0$. Then
$$
|f(x)|=\operatorname{Re}f(\alpha x)=U(\alpha x)\leq p(\alpha x)=p(x).
$$
The bound is immediate if $f(x)=0$. This proves the real and complex <seminorm> forms of <Hahn-Banach theorem> without invoking any version of that theorem.
For the <distance to a set> assertion, let $d=d(z,Y)>0$. Define on $Y+\mathbb Kz$ the <linear functional> $g(y+\alpha z)=\alpha d$. The decomposition is unique because $z\notin Y$, and for $\alpha\ne0$,
$$
|g(y+\alpha z)|=|\alpha|d
\leq|\alpha|\|z+y/\alpha\|=\|y+\alpha z\|.
$$
For $\alpha=0$ the bound is immediate. Apply the just-proved <Hahn-Banach theorem> with $p(x)=\|x\|$ to get $f\in X^*$, $f|_Y=0$, $f(z)=d$, and $\|f\|\leq1$. Choose $y_n\in Y$ with $\|z-y_n\|\to d$. Since
$$
d=|f(z-y_n)|\leq\|f\|\|z-y_n\|,
$$
and $d>0$, passage to the limit gives $\|f\|\geq1$. Thus \b[$\boxed{f|_Y=0,\quad f(z)=d(z,Y),\quad\|f\|=1}$]. This version of the <Hahn-Banach distance formula> needs neither a closed $Y$ nor an attained <distance to a set>.
For the <Banach limit> construction below, take the real <l-infinity sequence space>, let $S(x_1,x_2,\ldots)=(x_2,x_3,\ldots)$ be its <left shift on bounded sequences>, put $Y=(I-S)\ell^\infty$, and let $\mathbf1=(1,1,\ldots)$. This is a <vector subspace> and its <distance to a set> from $\mathbf1$ determines the construction.
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