Let $X$ be a real normed space and $A \subset X$ a nonempty subset. We define the closed convex hull of $A$ to be the subset $$\overline{\text{conv}}(A):=\bigcap \{B\subset X:A\subset B, B \text{ is closed and convex}\}.$$ $\overline{\text{conv}}(A)$ is a closed convex subset of $X$ with the property that if $C \subset X$ is closed and convex, and $A \subset C$, then $\overline{\text{conv}}(A) \subset C$.
Show that $$\overline{\text{conv}}(A) = \{x \in X : \text{ for all } f \in X', f(x) \leq \sup_{a\in A} f(a)\},$$ where $X'$ is the dual of $X$.
I first tried proving that this set in convex, which was pretty straightforward.
Let $x,y\in \overline{\text{conv}}(A)$, then for $t\in [0,1]$ we have that $tx+(1-t)y\in X$.
In particular for all $f\in X'$ we have that $f(tx+(1-t)y)=tf(x)+(1-t)f(y)\leq t\sup_{a\in A}f(a)+(1-t)\sup_{a\in A}f(a)=\sup_{a\in A}f(a)$.
Thus $tx+(1-t)y\in \overline{\text{conv}}(A)$.
Showing that the set is closed is shown in the comments.
Lastly I'm really lost on how to prove the last property, that is: if $C \subset X$ is closed and convex, and $A \subset C$, then $\overline{\text{conv}}(A) \subset C$.
Any help is greatly appreciated!