This is a question about Theorem 7.31 in Baby Rudin.
(Baby Rudin) Theorerm 7.31 Suppose $\mathscr A$ is an algebra of functions on a set $E$, $\mathscr A$ separates points on $E$, and $\mathscr A$ vanishes at no points of $E$. Suppose $x_1, x_2\in E$ are distinct and $c_1, c_2$ are real numbers, then there is $f\in \mathscr A$ so that $f(x_1) = c_1, f(x_2) = c_2$.
By "vanishes at no points", it means for all $x\in E$, there is $f\in E$ so that $f(x) \neq 0$.
It seems that that condition is not necessary: as long as the algebra $\mathscr A$ separates points on $E$, for all $x_1, x_2$ distinct there is a function $g\in \mathscr A$ such that $g(x_1)\neq g(x_2)$.
Then we can construct a function $$f(x) = \frac{c_1(g(x) - g(x_1))-c_2(g(x)-g(x_2))}{g(x_2)-g(x_1)}$$
which satisfies the desired properties : $$f(x_1)=c_1 \ \ \ \ f(x_2)=c_2.$$
It seems to me that "algebra $\mathscr A$ vanishes at no point of $E$" plays no role in this proof.
Is that condition really needed in this proof?
Your function $f$ might not be in $\mathscr A$. In order that your $f$ is in $\mathscr A$, the constant function has to be in $\mathscr A$ since you used
$$ \frac{g(x_1)}{g(x_2) - g(x_1)}.$$
Indeed, if constant functions were in $\mathscr A$, the algebra vanishes at no points of $E$ trivially.