projection maps from product space are open If $X_i$ is a family of topological spaces with $i\in I$, and $X=\prod_{i\in I} X_i$ is product topological space then the maps $\pi_k:X\rightarrow X_k$ are open. 
To prove this, we consider subbasic open sets of product topology; then the image of them under $\pi_k$ is either whole space $X_k$ or an open set in $X_k$ since subbasic open set in $X$ is of the form $\prod_{i} U_i $ where $U_i$'s are open in $X_i$ and  all $U_i$ except one equals $X_i$.
Is this justification correct for proving openness of projection maps?
The usual justification in books is somewhat little bit lengthy than the above one (if true). 
 A: It is a good start but not enough. You should add the following.
If $\mathcal V$ denotes a base of a topology on $X$ and $f:X\to Y$ is a function then $f$ will be open if $f(V)$ is open for every $V\in\mathcal V$. 
This because an open set $U$ can be written as a union of elements of $\mathcal V$ and function $f$ respects unions in the sense that $f(\bigcup_{\lambda\in\Lambda}V_\lambda)=\bigcup_{\lambda\in\Lambda}f(V_{\lambda})$.
So if $V_\lambda\in\mathcal V$ for every $\lambda\in\Lambda$ then $f(U)=f(\bigcup_{\lambda\in\Lambda}V_\lambda)=\bigcup_{\lambda\in\Lambda}f(V_{\lambda})$ is a union of open sets, hence is open.
A: I would say that it is correct but not complete. The sets to which you call “subbasic open sets” are a subbase of the topology of $\prod_{i\in I}X_i$. You proved correctly that, for each $k\in I$ and each basic open set $A$, $\pi_k(A)$ is an open set. More generally, if a subset $A$ of $X$ is of the form $\prod_iU_i$ where all $U_i$'s are open and all but finitely many of them are equal to $X_i$, then $\pi_k(A)$ is open. And the open sets that I have described for a base of the topology of $X$. All that remains is to prove the following statement:
Lemma: If $X$ and $Y$ are topological spaces, if $\mathcal B$ is a base of the topology of $X$ and if $f\colon X\longrightarrow Y$ is a map such that, for each $A\in\mathcal B$, $f(A)$ is open, then $f$ is an open map.
Note: This answer was edited due to a comment made by Henno Brandsma.
