Let $X$ be a real, normed space,and $I$ a non-empty index set. Let $ \{x_i : i\in I\} \subset X$ and $\{ \alpha_i : i\in I\} \subset \mathbb{R}_{\geq 0}$, $R > 0$. For $$ S := \{ f\in X^* : ||f||\leq R \text{ and } |f(x_i)|\leq \alpha_i \text{ for each }i\in I \},$$ I want to show, for arbitrary $x_0\in X$ that the set $$\{ f(x_0) : f\in S\}$$ is a closed and bounded interval on $\mathbb{R}$.
To do this, I tried to show $S$ is compact and convex as a subset of the continuous dual $X^*$, since then the result will follow by using Conway, Proposition V.7.9., which says that the image under a continuous affine map of a convex compact subest of a locally convex space is compact and convex.
I'm using Conway's definition for compactness, and trying to show compactness by showing each net in $S$ clusters, but I'm still stuggling with proofs regarding nets and I can't see how I should define the cluster point. I'm thus also unsure if this is the easiest way to go about this proof? Here there is an alternative way of solving this, but this is slightly different.
A special case of what you are trying to prove occurs whenever $\alpha_i=R\|x_i\|$ for each $i$, in which case $S$ is just the ball of radius $R$ in $X^*$. Therefore, trying to directly prove $S$ is weak-* compact amounts to at least (re?)-proving Banach Alaoglu.
If you already have Banach Alaoglu then you may as well use it, as they do at the link, since $S$, as a closed subset of the ball, must then also be weak-* compact.
Otherwise, anything you do will at least reprove that theorem.