Let $C$ be a convex closed cone in $\mathbb{R}^n$. A face of $C$ is a convex sub-cone $F$ satisfying that whenever $\lambda x + (1-\lambda)y\in F$ for some $\lambda \in (0,1)$ and $x,y\in C$, it holds that $x,y\in F$.
Is it true that all faces of $C$ are closed?
Answer: See the comment by @daw.
HINT:
Let $F$ be a face of the cone $C$, and let $V$ be the subspace generated by $F$. Then $F = V\cap C$.
Indeed, let $(v_i)_{i\in I}$ in $F$ a basis of $V$. Let $v \in V\cap C$. The system $\{(v_i)_{i\in I},\, v\}$ is linearly dependent, so some nontrivial linear combination equals $0$. Separating positive and negative coefficients, we may assume that $$\sum_{i=1}^k \alpha_i v_i = \sum_{i=k+1}^m \beta_i v_i + \beta v$$ with $\alpha_i$, $\beta_i$, $\beta>0$. The LHS/RHS are in $F$. We conclude ( F is a face) that $v \in F$.
An analogous statement works for faces of convex sets: every face $F$is the intersection of the affine subspace $V$ generated by it and the convex set $C$ : $F= V\cap C$.
Note that if $F$ is a face of $C$, then both $F$ and $C\backslash F$ are convex (cones). The converse is not true (say $F$ is half of a circular cone). But if $C\backslash F$ is a convex cone and $F = V\cap C$ for a subspace $V$, then $F$ is a face.