"If $(X_i,\tau_i)_{i\in\mathbb{N}}$, are compact subspaces of $[0,1]$, deduce from Theorem 9.3.8 and Exercises 9.2 #1, that $\Pi(X_i,\tau_i)$ is compact."
Theorem 9.3.8 is the hilbert cube is compact. Exercise 9.2 #1 is that the infinite product of closed sets is a closed set in the product topology.
So every compact subspace of $[0,1]$ are simply closed intervals in $[0,1]$ and closed intervals are all homeomorphic to each other. So we have an infinite product where each factor is homeomorphic to $[0,1]$ which is by definition the Hilbert Cube so that is a compact space.
Now I did not had to use the fact that the infinite product of closed sets is a closed set in the product topology, which I think I'm suppose to use. Is there a mistake in my logic?
Best Answer
As Kavi pointed out in the comments, it is not true that all compact subsets of $[0,1]$ are closed intervals.
What you can use is that since $(X_i,\tau_i)$ are compact subspaces of the compact set $[0,1]$, they are closed. So, by Exercise 9.2 #1, $\Pi_i(X_i,\tau_i)$ is closed. Then, by compactess of the Hilbert cube, any closed subset is also compact.
By the way, it is true that any product of compact topological spaces is compact in the product topology - this is called Tychonoff's Theorem.