If $X:= \prod X_\alpha$ is locally compact $\Rightarrow$ each $X_\alpha$ is locally compact and $X_\alpha$ is compact for all but finitely many values of $\alpha$
My solution So I will use that if $f:X\rightarrow Y$ is open, continuous map and we know that $X$ is locally compact then $Y$ is locally compact.
Define projection functions $p_\alpha : X \rightarrow X_{\alpha}$, since each $p_i$ is an open continuous map $\Rightarrow$ each $X_\alpha$ is as well locally compact.
For the second part of the proof I will again use projection functions. So we know that since the product is locally compact, for each point $x \in \prod X_\alpha$ and an open neighborhood $U$ of that point there exists compact subspace $C$ of the product such that $x \in U \in C$.And $C$ has all but finitely many projections equal to the whole corresponding space.
Since projections are continuous this spaces must be compact.
Is my proof correct or it is not enough?
Projections are onto, continuous and open, so it is immediate then that $X_\alpha$ is locally compact $\forall \alpha$. To prove the second part, let $x \in \prod X_\alpha$ and let $U$ be a compact neighborhood of $x$. Then $U$ contains a basic neighborhood of the form:
$$p_{\alpha_1}^{-1}(U_{\alpha_1})\cap p_{\alpha_2}^{-1}(U_{\alpha_2})\cap \dots\cap p_{\alpha_n}^{-1}(U_{\alpha_n})$$
Because $\prod X_\alpha$ is indeed equipped with the product topology, and it follows that $p_{\alpha}(U) = X_\alpha \ , \ \forall \alpha \not\in \{\alpha_1,\dots,\alpha_n\}$. Since the continuous images of compact subsets are also compact, each $X_\alpha$ is compact at least $\forall \alpha \not\in \{\alpha_1,\dots,\alpha_n\}$