Show that $\overline{со}(A) = \overline{co(A)}$. I want to prove Corollary. 1.13. of Conway's book on Functional Analysis :

If $A \subset X$, then "the intersection of all closed convex sets that contain $A$" $=$ "the closure of the intersection of all convex sets that contain $A$".

If $x$ belongs to the closure of the intersection of all convex sets that contain $A$, then either $x$ belongs to the intersection (and thus all of the intersecting sets) or belongs to the limit points of the intersection (and thus intersects all the intersecting sets). Taking closure as the smallest closed sets, roughly speaking I think $x$ belongs to the intersection all convex closed sets intersecting. The other direction seems not reachable.
 A: Let $F =\{B \supseteq A: B \text{ is closed convex set}  \}$ and $G =\{C \supseteq A: C \text{ is convex set}  \}$. We want to prove that
$$ \bigcap_{B \in F}B = \overline{\bigcap_{C \in G}C}$$
Proof:  Note that $\bigcap_{B \in F}B$ is convex and $A \subseteq \bigcap_{B \in F}B$. So, $\bigcap_{B \in F}B\in G$. So
$$ \bigcap_{C \in G}C \subseteq \bigcap_{B \in F}B$$
But $\bigcap_{B \in F}B$ is also closed (because it is the intersection of closed sets). So,
$$ \overline{\bigcap_{C \in G}C} \subseteq \bigcap_{B \in F}B \tag{1}$$
On the other hand, note that $\bigcap_{C \in G}C$ is convex, so, $\overline{\bigcap_{C \in G}C}$ is closed convex (see Remark). Since $A \subseteq \overline{\bigcap_{C \in G}C}$, we have that $\overline{\bigcap_{C \in G}C} \in F$. So,
$$ \bigcap_{B \in F}B \subseteq  \overline{\bigcap_{C \in G}C} \tag{2}$$
From $(1)$ and $(2)$, we have
$$ \bigcap_{B \in F}B = \overline{\bigcap_{C \in G}C}$$
Remark: It is a well-known result that if $S$ is convex, then the closure of $S$ (that is $\overline{S}$) is convex as well.  See, for instance, here.
