Theorem A nonempty polytope $P \subseteq \mathbb{R}^n$ is integral if and only if $\max\{c^Tx: x\in P\}$ is integer for every integral $c \in \mathbb{R}^n$.
The independent-set polytope of a graph $G=(V,E)$ is defined as \begin{equation} P_{\mathrm{ind}}(G) = \mathrm{conv}\{\chi_I \in \mathbb{R}^V : I \subseteq V \ \mathrm{is \, independent \, in \,} G\}. \end{equation} The clique inequality polytope is defined as \begin{equation} P_{\mathrm{cl.ineq}}(G) = \{x \in \mathbb{R}_+^n : x(C) \leq 1 \, \mathrm{for \, every \, clique\,} C \, \mathrm{of} \, G\}. \end{equation}
I need to show that if $G$ is perfect, then $P_{\mathrm{ind}}(G) = P_{\mathrm{cl.ineq}}(G)$. The proof is split up into 3 exercises. I have already shown the first two:
Exercise 1 If $G$ is perfect, then $\alpha(G) = \max\{\mathbf{1}^Tx :x\in P_{\mathrm{cl.ineq}}(G)\}$.
Exercise 2 Let $G$ be a perfect graph and fix a vertex $u$ of $G$. Now split $u$: create a copy of $u$ and connect it to all neighbors of $u$; call the resulting graph $G'$. Show that $G'$ is perfect.
What I don't see is how to combine these results to proof the last part:
Exercise 3 Use the theorem to show that $P_{\mathrm{cl.ineq}}(G)$ is integral if $G$ is perfect.
From this it indeed follows that $P_{\mathrm{ind}}(G) = P_{\mathrm{cl.ineq}}(G)$.
Now assume $G$ is perfect. I need to show that $\max\{c^T x: x \in P_{\mathrm{cl.ineq}}(G)\} \in \mathbb{Z}$ for all integral $c \in \mathbb{R}^n$. If $c = \mathbf{1}$, then from exercise 1 we see that it is satisfied. I have the feeling that I have to apply multiple splittings of nodes of $G$, but I just don't see how it all connects. Any hint would be greatly appreciated.
Suppose first that each component of $c$ is a positive integer. Then consider the graph $G'$ where the $i^{\text{th}}$ vertex is split into $c_i$ copies. Since we can accomplish this by repeatedly splitting vertices, $G'$ is perfect by multiple applications of exercise 2.
If you argue that $$\max\{c^{\mathsf T}x : x \in P_{\text{cl.ineq}}(G)\} = \max\{1^{\mathsf T}y : y \in P_{\text{cl.ineq}}(G')\}$$ then you can use exercise 1 to show that the hypotheses of the theorem are met.
If $c_i \le 0$ for some $i$, then at least one $x \in P_{\text{cl.ineq}}(G)$ which maximizes $c^{\mathsf T}x$ will have $x_i = 0$, so in this case we can delete the $i^{\text{th}}$ vertex of $G$. Do this for all $i$ with $c_i \le 0$, and we have reduced to the previous case (where every component of $c$ is a positive integer). Note that the result of deleting some vertices is an induced subgraph of $G$, which is still perfect. (You should prove this if you have not done so already, but it's not hard.)