Showing a infinite product is compact if each factor is homeomorphic to Sierpinski space "Using (i) and Proposition 9.2.5, show that if $(X_i , T_i)$, for each $i \in N$, is homeomorphic to the Sierpinski Space, then $\prod_i(X_i , T_i)$ is compact."
i) There is a continuous image from $[0,1]$ to the Sierpinski space.
Proposition 9.2.5) If for each $i,$ $f_i$ is a continuous mapping from $X_i$ to $Y_i$, then $f$ defined by $f=(f_1,f_2,f_2...)$ is continuous.
We have not proven Tychonoff's theorem otherwise this would be trivial. We can use the fact that the Cantor space is compact and the Cantor Space is homeomorphic to infinite products of itself and $[0,1]$ is a continuous image of the Cantor space, so I just need to make sure I am connecting the dots correctly.
So if each factor $X_i$ is homeomorphic to the Sierpinski space, there is a continuous map from $[0,1]$ to each factor. Now from proposition 9.2.5, $f$ is continuous each $f_1$ is continuous and each $[0,1]$ is homeomorphic to the cantor space which so the product topology of $[0,1]$ is homeomorphic to the product topology of the cantor space $[0,1]$ and the cantor space is homeomorphic to any number of cartesian product of itself. The Cantor space is compact, and $f$ is a continuous mapping so compactness is preserved thus $\prod_i(X_i , T_i)$ is compact.
 A: The Cantor set is not needed. Every Sierpiński space is the continuous image of $[0,1]$, so combining these into a product map, we get that the product of Sierpiński spaces is the continuous image of the Hilbert cube, of which you know it is compact. So your product is too (continuous image of compact is compact). That’s all.
If you only know the Cantor cube is compact then use the trivial continuous maps from the discrete two-point space to Sierpiński  space instead.
A: For the record I’ll note that this is a rather roundabout way to prove the result, since there’s a very simple direct proof. For $n\in\Bbb N$ let $p_n$ be the non-isolated point in $X_n$, let $$p=\langle p_n:n\in\Bbb N\rangle\in X=\prod_{n\in\Bbb N}X_n\,,$$ and suppose that $U$ is an open nbhd of $p$. By the definition of the product topology there is a $B=\prod_{n\in\Bbb N}V_n$ such that $V_n$ is open in $X_n$ for each $n\in\Bbb N$, and $p\in B\subseteq U$. The only open set in $X_n$ that contains $p_n$ is $X_n$ itself, so $B=\prod_{n\in\Bbb N}X_n=X$, and therefore $U=X$. Thus, $X$ is a member of every open cover of $X$, which is therefore compact. (Note that this argument works for the box topology as well as for the product topology.)
