If $\prod^{\infty}_{i=1} (X_i, T_i)$ is locally compact, then each $(X_i, T_i)$ is locally compact and all but a finite number of $(X_i, T_i)$ are compact.
Let $X=\prod^{\infty}_{i=1} (X_i, T_i)$, then for each $\prod^{\infty}_{i=1} \{x_i\}$ there exists a compact neighborhood $\prod^{\infty}_{i=1} N_{x_i}$ of each $x_i$. Therefore each $x_i \in N_{x_i}$ where $N_{x_i}$ is a compact subset of $(X_i, T_i)$.
The second part is what is confusing me. It would seem that $[0,1]^{\Bbb N}$ is countably compact and each $[0,1]$ is compact, so there does not exist a finite number of not compact sets. Unless by a finite number that's including $0$?
The proof is not correct as it stands, because you don't know beforehand that the compact neighbourhood of $x = (x_i)$ is of the form $\prod_i N_{x_i}$.
So you do know that there exists a compact set $C \subseteq \prod_i X_i$ such that $(x_i)$ is in the interior of $C$ (i.e. $C$ is a neighbourhood of $x$).
The latter means that there is a basic open neighbourhood $O = \prod_i O_i$ of $x$, where all $O_i$ are open in $X_i$ and also $F = \{i: O_i \neq X_i \}$ is at most finite, such that $x \in \prod_i O_i \subseteq C$.
This means that for $i \notin F$, we have $O_i = X_i$ and so $X_i = \pi_i[O] \subseteq \pi_i[C] \subseteq X_i$, so $X_i = \pi_i[C]$, and the latter set is compact as $\pi_i$ (a continuous projection) preserves compactness. So for all $i \notin F$ (so for all but finitely many $i$) we have that $X_i$ is compact.
That $X_i$ is locally compact for all $i$ whenever $X = \prod_i X_i$ is follows as all $X_i$ are continuous open images of $X$, and such maps preserve local compactness in general.