a step in proof of Tychonoff theorem For simplicity, consider finite/countable collection of compact topological spaces $X_1,X_2,\cdots$. I am trying to understand the proof of the fact that the product $X=\prod_i X_i$ is compact, given in Conway's point set topology as below: 
(1) Let $\mathcal{O}=\{\cdots \times X_{i-1} \times U_i\times X_{i+1}\times \cdots| U_i \mbox{ open in } X_i\}$ be an open cover of $X$, in which, $U_i$ ranges over some  open subsets of $X_i$ for each $i$. 
(Conway calls this cover as a cover by open sets among a standard subbasis, i.e. only one component is proper open set; hope that this is clear; otherwise, see page 64 in mentioned book). 
(2) For every $i$, let $\mathcal{C}_i$ be the collection of those open sets of $X_i$ which appear in the open cover $\mathcal{O}$ of $X$.
(3) Claim: There is atleast one $i$ such that $\mathcal{C}_i$ is open cover of $X_i$, say $X_i=U_1\cup \cdots \cup U_r$ with $U_j\in \mathcal{C}_1$.
(4) Then $\cdots \times X_{i-1} \times U_j\times X_{i+1}\times \cdots$ with  $j=1,2,\cdots r$ is finite subcover of $X$.
Question: Is it not true that the statement of Claim is valid for every $i$ instead of at least one $i$?
My partial answer: Perhaps, Conway wants to do this: let $\mathcal{C}_i$ denote the collection of those proper open sets $U_i$ of $X_i$ which appear as $i$-th component in the open cover $\mathcal{O}$. Then, in this case, the claim is valid for some $i$ (as he proves) but it is not necessarily valid for every $i$.
 A: It's not too hard to make examples, even of two spaces, where this does not hold for some coordinate, so no, you cannot show it holds for all $i$. It only needs to hold for one $i$ to get a finite subcover in that coordinate which translates back to a finite subcover of subbasic open sets.
(e.g. take a cover of $X \times Y = \{1,2, 3\} \times \{1,2,3\}$ (both discrete) by subbasic
sets $\{\{1\} \times  Y, \{2\} \times Y, X \times \{1\} ,X \times \{2,3\}\}$ e.g. We get a cover in the second coordinate, not in the first.
It's easier (in general, for all index sets $I$) to denote the subbasic sets as $S(i,O) = (p_i)^{-1}[O]$, where the only non-trivial open set is in coordinate $i$ (namely $O \subseteq X_i$). We only need to consider the case $O \neq X_i$ because if $S(i, X_i)$ were in a cover of subbasic elements, even for one $i$, it would be a one subset subcover! BUt even that is unnecessary as a special case:
So if we have the standard subbase $\mathcal{S} = \{S(i,O): i \in I; O \subseteq X_i\}$ and we have a cover $\mathcal{U} \subseteq \mathcal{S}$ we can define for each $i \in I$: $$\mathcal{U}_i = \{O \subseteq X_i: S(i, O) \in \mathcal{U}\}$$
and we can show that for at least one $i$, $\mathcal{U}_i$ is a cover of $X_i$. 
(if one $\mathcal{U}_i$ had $X_i$ in it, it would be trivial, but it's nothing to bother ourselves with.)
In general this part of the proof uses the axiom of choice: suppose for a contradiction that $X_i \setminus \bigcup \mathcal{U}_i \neq \emptyset$ for all $i$. Then for all $i$ simultaneously pick $p_i \in X_i \setminus \bigcup \mathcal{U}_i$ (this uses choice) and the point $(p_i)_{i \in I}$ lies in some $S(j, O) \in \mathcal{U}$ for $j \in I$, as $\mathcal{U}$ is a cover. But then by definition $p_j \in O \in \mathcal{U}_j$ contrary to how $p_j$ was chosen. 
So for some $j \in I$, $\mathcal{U}_j$ is a cover of $X_j$ and by compactness of $X_j$ it has a finite subcover $\{O_1, \ldots O_n\} \subseteq \mathcal{U}_j$, which means that $\bigcup_{k=1}^n O_i = X_j$ and thus $$\bigcup_{k=1}^n S(j ,O_k) = \bigcup_{k=1}^n (p_j)^{-1}[O_k] = p_j^{-1}[X_j] = \prod_i X_i$$ and we have a finite subcover of $\mathcal{U}$, as required.
Note that we (in this proof path) use the axiom of choice at two points in the proof of Tychonoff: once to show Alexander's subbase theorem (which shows that covers by subbasic elements suffice) and in the above proof by contradiction that at least one $\mathcal{U}_i$ is a cover of its coordinate space.
