= Solution
The <Hahn-Banach separation theorem for two convex sets> states that disjoint nonempty <convex sets> $A,B$ in a real <locally convex space>, with $A$ open, admit a nonzero <continuous linear functional> $\ell$ and $\alpha\in\mathbb R$ such that
$$
\boxed{\ell(a)<\alpha\leq\ell(b)\quad(a\in A,\ b\in B).}
$$
There need not be a positive uniform gap between the two sets.
For a proof, set $C=A-B$. It is open and <convex>, and $0\notin C$. Choose $c_0\in C$, let $V=C-c_0$, and let $p$ be the <Minkowski functional> of $V$. Because $V$ is open, convex and contains zero, $p$ is a finite <sublinear functional>, and $V=\{x:p(x)<1\}$. Put $z=-c_0$. Since $z\notin V$, $p(z)\geq1$. On $\mathbb Rz$ define $\ell_0(tz)=tp(z)$. For $t\geq0$ this equals $p(tz)$; for $t<0$ it is nonpositive and hence at most $p(tz)$. The <Hahn-Banach theorem> extends it to a linear $\ell$ with $\ell\leq p$. Both $p(x)$ and $p(-x)$ are bounded near zero, so this domination makes $\ell$ continuous; it is nonzero because $\ell(z)=p(z)\geq1$. For $c\in C$,
$$
\ell(c)=\ell(c-c_0)-p(z)\leq p(c-c_0)-p(z)<1-p(z)\leq0.
$$
Thus $\ell(a)<\ell(b)$ for every $a\in A,b\in B$. Define $\alpha=\sup_{a\in A}\ell(a)$, which is finite because any fixed $b\in B$ bounds it above. Openness of $A$ and nonzeroness of $\ell$ ensure $\ell(a)<\alpha$ for every $a\in A$, and $\alpha\leq\ell(b)$ for every $b\in B$. This proves the stated separation.
For the sequence criterion, suppose $f\in S_{X^*}$ and $f(x_i)\geq\varrho$. If each $t_i\geq0$, then
$$
\left\|\sum_{i=1}^nt_ix_i\right\|\geq f\left(\sum_{i=1}^nt_ix_i\right)\geq\varrho\sum_{i=1}^nt_i.
$$
Conversely, the assumed inequality implies that every <convex combination> of the $x_i$ has norm at least $\varrho$. Hence $D=\{x:\|x\|<\varrho\}$ is disjoint from $C=\operatorname{conv}\{x_i:i\geq1\}$. Apply the separation theorem with the open set $D$ to obtain $\ell\ne0$ and
$$
\ell(c)\geq\sup_{d\in D}\ell(d)=\varrho\|\ell\|\quad(c\in C).
$$
Then $f=\ell/\|\ell\|$ has norm one and satisfies $f(x_i)\geq\varrho$ for every $i$. Therefore
$$
\boxed{\exists f\in S_{X^*},\ \forall i\ f(x_i)\geq\varrho\quad\Longleftrightarrow\quad
\left\|\sum_{i=1}^nt_ix_i\right\|\geq\varrho\sum_{i=1}^nt_i\ \text{for all }n\text{ and }t_i\geq0.}
$$
The PDF confirms the “if and only if” that is garbled in the TeX transcription.
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