Not looking for an outright solution but some back-and-forth.
My thoughts so far:
A face of any polytope $S$ can be defined as $F = H \cap S,$ where $H$ is some supporting hyperplane of $S$.
We know that every vertex in a $d-$dimensional simplex $S$ lies on $d$ unique edges. Given some subset $V'$, we'd like to show that conv$(V')$ is a face of $S.$ My idea for the proof is
1.) Show that aff$(V') \cap S^\circ = \{\}.$ The standard $d-$simplex with $d+1$ vertices is defined as $\{x \in \mathbb{R}^{d+1} : \mathbb{1}x = 1, x_i \geq 0\}.$ So, its interior would be $\{x \in \mathbb{R}^{d+1} : \mathbb{1}x = 1, x_i > 0\}.$ We know that aff$(V')$ is all affine combinations of the points in $V'.$ And aff$(V')$ is just the dim(aff($V'$)) $= q$ hyperplane spanned by $V'$, and all of these points have some $x_i = 0$. So, aff$(V') \cap S^\circ = \{\}$.
2.) Show there exists some supporting hyperplane $H$ such that conv$(V') = H \cap S.$ I suspect this $H$ is just the aforementioned aff$(V')$, which would pretty much complete the proof.
Does this direction seem sound? Am I on the right track, or am I missing something or making something trivial when it really isn't?
I am assuming that the vertex set $V$ has $d+1$ affinely independent points. Hence any $V' \subset V$ is also affinely independent. Let $C = \operatorname{co} V$.
Since $V$ is affinely independent, the barycentric coordinates of any point in $\operatorname{aff} C$ are unique. Let $\mu(x)$ denote the barycentric coordinates of $x \in \operatorname{aff} C$. That is, $x = \sum_{v \in V} \mu_v(x) v$, where $\sum_{v \in V} \mu_v(x) = 1$, and $\mu_v(x)$ is the '$v$ coordinate' of $x$ with respect to $V$.
A face $F$ of a convex set $C$ is a convex subset $F \subset C$ such that every closed line segment in $C$ whose relative interior intersects $F$ has both end points in $F$.
Suppose $V' \subset V$. It is clear that if $|V'| \le 1$, or $V=V'$ then $V'$ is a face. So, suppose $V'$ is not one of these cases.
Let $F=\operatorname{co} V'$. Suppose we have a line segment $[x,y] \subset C$ such that $(x,y) \cap F$ is non empty and let $z \in (x,y) \cap F$. Then $\mu_v(z) >0$ iff $v \in V'$. Let $\lambda \in (0,1)$ be such that $z = \lambda x + (1-\lambda) y$. By uniqueness, we have $\mu_v(z) = \lambda \mu_v(x) + (1-\lambda) \mu_v(y)$ for all $v \in V$, and if $w \in V \setminus V'$, we must have $\mu_w(x) = \mu_w(y) = 0$ (otherwise $\mu_w(z)>0$). In particular, $\mu_w(x) = \mu_w(y) = 0$ for all $w \in V \setminus V'$ and so $x,y \in F$. Hence $F$ is a face.