By scaling, we can assume WLOG that the perimeter of $P$ has length $2$. Let $X_1,...,X_n$ be the vertices of $P$, and let $x_i=|X_iX_{i+1}|$, where $X_{n+1}=X_1$; then $\sum_{i=1}^n x_i=2$. Since $P$ is a non-degenerate polygon, we have that $x_i<1$ for all $i=1,2,...,n$. To prove the claim it is sufficient to partition the cyclic sequence $(x_1,...,x_n)$ into three intervals such that the sum in each interval is strictly smaller that $1$; the endpoints of these intervals correspond to the selected vertices $A, B, C$ of $P$. To this end, we distinguish two cases: either there is some $p$ such that $\sum_{i=1}^p x_i=1$, or there is no such $p$. In the first case, the perimeter of $P$ can be broken into two polygonal chains of length $1$ each, both with endpoints $X_1$ and $X_{p+1}$. Since $x_i<1$ for all $i$, both these chains consist of at least two segments. Since $n>4$, we infer that the four segments incident to $X_1$ and $X_{p+1}$, namely $X_nX_1,X_1X_2,X_pX_{p+1}$ and $X_{p+1}X_{p+2}$, are pairwise different, and they do not constitute the whole perimeter. Hence $x_n+x_1+x_p+x_{p+1}<2$, so either $x_n+x_1<1$ or $x_p+x_{p+1}<1$. In the former subcase we can take intervals $(x_n,x_1)$, $(x_2,...,x_p)$ and $(x_{p+1},...,x_{n-1})$, and in each of them the sum is clearly smaller than $1$. Symmetrically, in the latter subcase we can take intervals $(x_p,x_{p+1}$, $x_{p+2},...,x_n$ and $(x_1,...,x_{p-1})$. We are left with the second case. Let $q$ be the largest index such that $\sum_{i=1}^q x_i < 1$. Since $x_1<1$ and $x_n<1$, we have that $1\leq q\leq n-2$. By the choice of $q$ and the fact that the first case was not applicable, we have that $\sum_{i=1}^{q+1} x_i>1$, hence $\sum_{i=q+2}^n x_i=2-\sum_{i=1}^{q+1} x_i <1$. As $x_{q+1}<1$, we can take intervals $(x_1,...,x_q)$, $(x_{q+1})$ and $(x_{q+2},...,x_n)$, and in each of them the sum is strictly smaller than $1$.