Assume without loss of generality that the 3-cycle satisfies $x_1 < x_2 < x_3$ with $F(x_1) = x_2$, $F(x_2) = x_3$, $F(x_3) = x_1$. An intermediate value argument as in the previous theorem gives a fixed point $z \in (x_2, x_3)$, handling the case $N = 1$. For $N \geq 2$, define \begin{align*} I_L = [x_1, x_2] \quad \text{and} \quad I_R = [x_2, x_3]. \end{align*} By continuity and the definitions of the 3-cycle, one checks: \begin{align*} F(I_L) \supset I_R \quad \text{and} \quad F(I_R) \supset I_L \cup I_R. \end{align*} In particular $I_R \subset F(I_R)$ and $I_L \subset F(I_R)$. Set $J_N = I_L$. Since $J_N = I_L \subset F(I_R)$, the Onto Closed Intervals lemma gives a closed interval $J_{N-1} \subset I_R$ with $F(J_{N-1}) = J_N$. Continuing inductively: $J_{N-1} \subset I_R \subset F(I_R)$, so the lemma gives $J_{N-2} \subset I_R$ with $F(J_{N-2}) = J_{N-1}$. After $N-2$ such steps, we obtain $J_1 \subset I_R$ with $F(J_k) = J_{k+1}$ for $k = 1, \ldots, N-2$. Since $J_1 \subset I_R \subset F(I_L)$, the lemma gives $J_0 \subset I_L$ with $F(J_0) = J_1$. By construction, $F^N(J_0) = J_N = I_L$, so there exist $a, b \in J_0$ with $F^N(a) = x_1$ and $F^N(b) = x_2$. Since $J_0 \subset I_L = [x_1, x_2]$, we have $a \geq x_1 = F^N(a)$ and $b \leq x_2 = F^N(b)$. The intermediate value theorem applied to $F^N(t) - t$ on $[a, b]$ gives $z \in [a,b] \subset J_0$ with $F^N(z) = z$. It remains to verify that $z$ is not a point of a smaller cycle. The orbit $z, F(z), \ldots, F^{N-1}(z)$ satisfies $z \in J_0 \subset I_L$ and $F^k(z) \in J_k \subset I_R$ for $k = 1, \ldots, N-1$. Since $I_L \cap I_R = \{x_2\}$, any repetition in the orbit would force $F^m(z) = x_2$ for some $m < N$. But $F^2(x_2) = F(x_3) = x_1 \in I_L$, which would remove $F^{m+2}(z)$ from $I_R$ — contradicting $F^{m+2}(z) \in J_{m+2} \subset I_R$ when $m+2 \leq N-1$. A careful case analysis on $m = N-1$ and $m = N-2$ in both scenarios yields a contradiction in each case. Hence the orbit of $z$ has exact period $N$.