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I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm.

Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets:

• $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$,

• $B$, the closed unit ball of $\ell_2$,

• $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e_0^\perp$.

• $D:=\overline{\text {co}}\big(B\cup( e_0+C)\cup(-e_0-C)\big)$.

Clearly, $C:=\big\{\sum_{k\ge1}2^{-k}a_ke_k: a_k \ge0 \text{ for all } k\ge1 \text{ and } \sum_{k\ge1}a_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $

The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell_2$, so it is the closed unit ball in an equivalent norm of $\ell_2$.

We have $D\cup (e_0+e_0^\perp)=e_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e_0+e_0 ^\perp$ is exactly the set $e_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e_0+ e_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e_0+ e_0 ^\perp) =e_0+\text{co}A\subsetneq D \cap (e_0+ e_0 ^\perp) $.

I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm.

Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets:

• $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$,

• $B$, the closed unit ball of $\ell_2$,

• $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e_0^\perp$.

• $D:=\overline{\text {co}}\big(B\cup( e_0+C)\cup(-e_0-C)\big)$.

Clearly, $C:=\big\{\sum_{k\ge1}2^{-k}a_ke_k: a_k \ge0 \text{ for all } k\ge1 \text{ and } \sum_{k\ge1}a_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $

The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell_2$, so it is the closed unit ball in an equivalent norm of $\ell_2$.

We have $D\cap (e_0+e_0^\perp)=e_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e_0+e_0 ^\perp$ is exactly the set $e_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e_0+ e_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e_0+ e_0 ^\perp) =e_0+\text{co}A\subsetneq D \cap (e_0+ e_0 ^\perp) $.

I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm.

Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets:

• $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$,

• $B$, the closed unit ball of $\ell_2$,

• $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e_0^\perp$.

• $D:=\overline{\text {co}}\big(B\cup( e_0+C)\cup(-e_0-C)\big)$.

Clearly, $C:=\big\{\sum_{k\ge1}2^{-k}a_ke_k: a_k \ge0 \text{ for all } k\ge1 \text{ and } \sum_{k\ge1}a_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $

The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell_2$, so it is the closed unit ball in an equivalent norm of $\ell_2$.

We have $D\cup (e_0+e_0^\perp)=e_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e_0+e_0 ^\perp$ is exactly the set $e_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e_0+ e_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e_0+ e_0 ^\perp) =e_0+A\subsetneq D \cap (e_0+ e_0 ^\perp) $.

I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm.

Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets:

• $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$,

• $B$, the closed unit ball of $\ell_2$,

• $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e_0^\perp$.

• $D:=\overline{\text {co}}\big(B\cup( e_0+C)\cup(-e_0-C)\big)$.

Clearly, $C:=\big\{\sum_{k\ge1}2^{-k}a_ke_k: a_k \ge0 \text{ for all } k\ge1 \text{ and } \sum_{k\ge1}a_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $

The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell_2$, so it is the closed unit ball in an equivalent norm of $\ell_2$.

We have $D\cup (e_0+e_0^\perp)=e_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e_0+e_0 ^\perp$ is exactly the set $e_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e_0+ e_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e_0+ e_0 ^\perp) =e_0+\text{co}A\subsetneq D \cap (e_0+ e_0 ^\perp) $.

I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm.

Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets:

• $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$,

• $B$, the closed unit ball of $\ell_2$,

• $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e_0^\perp$.

• $D:=\bigcup_{|b|+|c|=1} (bB+cC)$.

Clearly, $C:=\big\{\sum_{k\ge1}2^{-k}a_ke_k: a_k \ge0 \text{ for all } k\ge1 \text{ and } \sum_{k\ge1}a_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $

The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell_2$, so it is the closed unit ball in an equivalent norm of $\ell_2$. We have $D\cup (e_0+e_0^\perp)=e_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e_0+e_0 ^\perp$ is exactly the set $e_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e_0+ e_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e_0+ e_0 ^\perp) =e_0+A\subsetneq D \cap (e_0+ e_0 ^\perp) $.

I'd try $X:=\ell_2$ with an equivalent but non strictly convex norm.

Let $(e_k)_{k\ge0}$ be the standard Hilbert basis of $\ell_2$. Consider the sets:

• $A:=\{0\}\cup\{2^{-k}e_k:k\ge 1\}$,

• $B$, the closed unit ball of $\ell_2$,

• $C:=\overline{\text{co}}\, A$, a compact subset of the hyperplane $e_0^\perp$.

• $D:=\overline{\text {co}}\big(B\cup( e_0+C)\cup(-e_0-C)\big)$.

Clearly, $C:=\big\{\sum_{k\ge1}2^{-k}a_ke_k: a_k \ge0 \text{ for all } k\ge1 \text{ and } \sum_{k\ge1}a_k\le1\big\},$ so $\text{ext}\, C=A$ and ${\text{co}}(\text{ext}\, C)={\text{co}}\, A\subsetneq C. $

The set $D$ is a bounded, closed, convex, symmetric nbd of $0$ in $\ell_2$, so it is the closed unit ball in an equivalent norm of $\ell_2$. 

We have $D\cup (e_0+e_0^\perp)=e_0+C;$ the set of extremal points of $D$ in the affine hyperplane $e_0+e_0 ^\perp$ is exactly the set $e_0+A$, which implies that ${\text{co}}(\text{ext} D)\subsetneq D$ because the trace of these sets on $e_0+ e_0 ^\perp$ are ${\text{co}}(\text{ext} D)\cap (e_0+ e_0 ^\perp) =e_0+A\subsetneq D \cap (e_0+ e_0 ^\perp) $.