Let $E$ be a Banach space, let $C$ be a closed bounded convex subset of $E$. For $x,y\in E$ denote $]x,y[=\{ \lambda x+(1-\lambda)y : \lambda\in ]0,1[ \}$ (in particular if $x=y$ this set is equal to $\{ x\}$), let also $[x,y]=\{ \lambda x+(1-\lambda)y : \lambda\in[0,1] \}$.
Definition 1 : A closed convex subset $F$ of $C$ is called a face if $\forall x\in F,y\in C,z\in C$ such that $x\in]y,z[$, $y\in F$ and $z\in F$. Denote by $\mathcal F(C)$ the set of faces of $C$.
This is not super standard since I require a face to be a closed subset, in the euclidean space it does not make any difference but in general there are non closed faces (see Example 1).
Definition 2 : Let $D$ be a convex set, the relative interior of $D$ is $\mathrm{ri}(D)=\{ x\in D:\mathrm{cone}(D-x)=\mathrm{span}(D-x) \}$.
In particular $\mathrm{ri} [x,y]=]x,y[$ is always true.
I am trying to prove (or strengthen the assumptions of) the following Theorem (which is a generalization of Theorem 18.2 of Rockafellar, Convex analysis which proves it in Euclidean space, the proof is hard to generalize because it uses the fact that we can augment dimension of separating space by one repeatedly to reach the desired dimension, but in general it is not the case).
Theorem 1 : Let $C$ be a non empty closed bounded convex set, then $\{ \mathrm{ri(F) : F\in\mathcal F(C)\setminus \{\emptyset \}}\}$ is a partition of $C$.
Below I present a sketch of proof that relies on two Lemmas, I am interested in any comment about the sketch of proof and Lemma 1. I cannot prove Lemma 2 yet so this question is really about that.
Here is my sketch of proof. First we show that for $F_1,F_2\in\mathcal F(C)$, $\mathrm{ri}(F_1)\cap \mathrm{ri}(F_2)\neq \emptyset$ implies that $F_1=F_2$. It is clear that $\mathrm{ri}(F_1)\cap \mathrm{ri}(F_2)\subseteq \mathrm{ri}(F_1)\cap F_2$ so if the first one is non empty then we can apply Lemma 1 to obtain $F_1\subseteq F_2$ and similarly we get $F_2\subseteq F_2$. Next we can use the following definition, to aid in the proof that any $x\in C$ is contained in the relative interior of some face.
Definition : Let $X\subseteq C$, denote by $F_C(X)$ the smallest face containing $D$, for $x\in C$, denote by $F_C(x)\triangleq F_C(\{x\})$.
This is well defined since $\mathcal F(C)$ is closed under any intersection so we can intersect all faces that contains $D$ to obtain the smalest such face.
If we can prove that $\forall x\in C$, $x\in\mathrm{ri}(F_C(x))$ then we are done, this would also show from the first part of the proof that $\mathcal F(C)=\{ F_C(x) : x\in C \}\cup \{ \emptyset\}$. Therefore Lemma 2 (if true) is enough to finish the proof, indeed if $x\notin\mathrm{ri}(F_C(x))$, then we have a continuous linear $f:E\to\mathbb R$ such that $\forall y\in \mathrm{ri}(F_C(x))$, $f(y)<f(x)$, then $F=F_C(x)\cap f^{-1}(f(x))$ is a face containing $x$ that is a strict subset of $F_C(x)$ which is a contradiction.
Lemma 1 : $F\in\mathcal F(C)$ if and only if forall closed convex $D\subset C$ with $\mathrm{ri}(D)\cap F\neq \emptyset$, $D\subseteq F$.
Proof : The if part is obvious since $[x,z]$ is always a closed convex subset of $C$ when $y,z\in C$, and $\mathrm{ri}[y,z]=]y,z[$. let $D\subseteq C$ be closed convex such that there is $x\in \mathrm{ri} D\cap F$, let also $y\in D$, if $y=x$, then $y\in F$, suppose that $y\neq x$, then $x-y\in\mathrm{span}\{y-x \}\subseteq \mathrm{span} (D-x)=\mathrm{cone} (D-x)$, so there is $z\in D$ and $\lambda\in ]0,\infty[$ such that $x-y=\lambda (z-x)$ and so $x=\frac{1}{1+\lambda} y+\frac{\lambda}{1+\lambda} z\in]y,z[$, therefore $y\in F$. Therefore $D\subseteq F$.$$\tag*{$\blacksquare$}$$
The following Lemma is related to a question I posed earlier this week (Geometrical Hahn-Banach separation for $A$ and $B$ disjoint and $A$ relatively open.). The question was fairly poorly received by the community, I am guessing it's because of the lack of context, this is why I decided to ask this more detailed question.
Lemma 2 : If $D$ is a non empty closed convex set, and $x\in E$ is such that $x\notin \mathrm{ri}(D)$, there is a continuous linear function $f:E\to\mathbb R$ such that for all $y\in \mathrm{ri}(D)$, $f(y) < f(x)$.
Attempt proof : In the following we use two version of the Hahn-Banach Theorem that can be found on the wikipedia page If $x\notin D$, then by Hahn-Banach, since $\{x\}$ is compact convex and $D$ is closed convex we get the desired separation, otherwise $x$ is in the relative boundary of $D$, from this we get that $\mathrm{span}(D-x)$ is a topological vector space that we can endow with the subspace topology, in that topology $\mathrm{Int}(D-x)$ is non empty and $0\notin \mathrm{Int}(D-x)$ therefore we can apply a version of Hahn-Banach (first one on the link above) to get the desired functional. The reason why I think that $\mathrm{Int}(D-x)$ is non empty and $0\notin \mathrm{Int}(D-x)$ is because I believe that in that topology $\mathrm{ri}(D-x)=\mathrm{Int}(D-x)$, and the relative interior of a non empty closed set is non empty, but I have difficulties proving both those statements.
I have also tried a different approach on which I wrote something here, if the property mentioned there is true, then it is easy to show that if $x$ is in the relative boundary of $D$, then $x$ is strictly separated from $\mathrm{ri}(D)$ by a continuous linear functional.
Example 1 : Let $C$ be the set of all probability distribution over $\mathbb N$, as a metric we use $\| p-q \|_{\mathrm{TV}}=\sup_{A\subseteq \mathbb N} |p(A)-q(A)|$, let $f$ be the expectation, that is $\forall p\in C$, $f(p)=\sum_{i\in\mathbb N} i p(i)$. Let $F=\{ p\in C : f(p) < \infty \}$, then $F$ satisfies the property of Definition 1 but is not closed. Indeed if $x\in F$ and $y,z\in C$ with $x\in]y,z[$, then there is $\lambda\in ]0,1[$ such that $x=\lambda y+(1-\lambda)z$ and by linearity $f(x)=\lambda f(y)+(1-\lambda)f(z)$ which forces $f(y)<\infty$ and $f(z)<\infty$. Now it is not closed since for any $x\in C\setminus F$, we can build a sequence of distribution in $F$ : $x_n(i)=\frac{\mathbb 1(i\leq n) x(i)}{\sum_{i=1}^n x(i)}$ that converges to $x$. So the closure of $F$ is $C$.