By scaling, we can assume w.l.o.g. that the perimeter of $P$ has length $2$. Let $X_1, \dots, X_n$ be the vertices of $P$, and let $x_i = |X_i X_{i+1}|$, 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 $i = 1, 2, \dots, n$. To prove the claim it is sufficient to partition the cyclic sequence $(x_1, \dots, x_n)$ into three intervals such that the sum in each interval is strictly smaller than $1$; 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 $X_n X_1$, $X_1 X_2$, $X_p X_{p+1}$ 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 $(x_n, x_1)$, $(x_2, \dots, x_p)$ and $(x_{p+1}, \dots, x_{n-1})$, and in each of them the sum is clearly smaller than $1$. Symmetrically, in the latter subcase we can take intervals $(x_p, x_{p+1})$, $(x_{p+2}, \dots, x_n)$, and $(x_1, \dots, x_{p-1})$.

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 $1 \le q \le n-2$. 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 $(x_1, \dots, x_q)$, $(x_{q+1})$, and $(x_{q+2}, \dots, x_n)$, and in each of them the sum is strictly smaller than $1$. $\square$