Indeed, this can be proved more simply, and in greater generality -- assuming only that the support of $P$ is contained in $C$ (rather than in $\mathcal X$).

Indeed, there is a nonzero linear functional $f$ such that $f(y)\le f(x)$ for all $y\in C$. Then the support of the pushforward $f_\# P$ of $P$ under $f$ is contained in the interval $(-\infty,f(x)]$ whereas the mean of $f_\# P$ is $f(x)$. So, the support of $f_\# P$ is $\{f(x)\}$; that is, the support of $P$ is contained in $C_{x,f}:=C\cap f^{-1}(\{f(x)\})$, and the set $C_{x,f}$ is contained in the relative boundary of $C$. $\quad\Box$

Since this proof is so simple (and your lemma is, at least intuitively, quite obvious), perhaps it does not exist as a separate statement in the literature.

Indeed, this can be proved more simply, and in greater generality -- assuming only that the support of $P$ is contained in $C$ (rather than in $\mathcal X$).

Indeed, without loss of generality the affine hull of $C$ is $\Bbb R^n$. Then there is a nonzero linear functional $f$ such that $f(y)\le f(x)$ for all $y\in C$. Then the support of the pushforward $f_\# P$ of $P$ under $f$ is contained in the interval $(-\infty,f(x)]$ whereas the mean of $f_\# P$ is $f(x)$. So, the support of $f_\# P$ is $\{f(x)\}$; that is, the support of $P$ is contained in $C_{x,f}:=C\cap f^{-1}(\{f(x)\})$, and the set $C_{x,f}$ is contained in the relative boundary of $C$. $\quad\Box$

Since this proof is so simple (and your lemma is, at least intuitively, quite obvious), perhaps it does not exist as a separate statement in the literature.