Let $X$ be a real normed space, let $\beta>0$, and for some set $I$, $\{x_i\}_{i\in I}\subset X,\{\alpha_i\}_{i\in I}\subset[0,\infty)$. Then set $$A=\{f\in X^*:\|f\|\leq\beta,|f(x_i)|\leq\alpha_i\ \forall i\in I\},$$ and set $B=\{f(x_0):f\in A\}$. I want to prove that $B$ is a closed bounded interval of the real line.
Using that each $f\in A$ is norm-bounded by $\beta$, it is easy to argue that $B$ is bounded. It is easy to show that $A$ is convex, and from there it follows that $B$ is convex. Hence, being a subset of the real line, it follows that $B$ is a bounded interval.
The part I'm having troubles with is showing that $B$ is closed. It seems that $A$ is closed, and therefore weakly closed, being convex. Furthermore, we observe that $B=\phi(A)$, where $\phi:X^{*}\to\mathbb R:f\mapsto f(x_0)$ is clearly continuous, so $\phi\in X^{**}$. Hence $B$ is the continuous image of a weakly closed set. I don't know if this brings us any further. Other approaches I tried didn't seem helpful. Any help on the `closed' part is much appreciated.
By Banach Alaoglu Theorem $A$ is (a weak* closed subset of a) weak* compact (set) and $B$ is a continuous image of this set.