Did I use axiom of choice in my proof? I have two different affine open covers for a scheme $X$, say $X = \cup_{i \in I} U_i$ and $X = \cup_{j \in J} V_j$. For each $p \in X$, we know there exist some $i(p)$ and $j(p)$ such that $p \in U_{i(p)}$ and $p \in V_{j(p)}$. We can then find an affine open set denoted $W_p$ which satisfies
$$
p \in W_p \subseteq U_{i(p)} \cap V_{j(p)}.
$$ 
This gives an open cover $X = \cup W_p$ such that each $W_p$ is affine and contained in one of the $U_i$'s and one of the $V_j$'s simultaneously.
I wasn't really sure if I had implicitly used axiom of choice or not.
I would appreciate any clarification. Thank you!
ps also is it acceptable to assume axiom of choice when I am doing an exercise from algebraic geometry? or should I try to avoid it? (I was wondering what the general convention was...)

Since I am quite unfamiliar with schemes, I'll follow a hunch.If $i$ and $j$ are such that $p\in U_i\cap V_j$ then there is some affine open set $W\subseteq U_i\cap V_j$ such that $p\in W$. Or at least, for every $p$ there exists at least one pair of $i$ and $j$ such that the above is true.

In its current form, the proof suggests choosing $i(p)$ and $j(p)$, and in that case the axiom of choice is used, unless you can somehow prove that $I$ and $J$ are well-ordered somehow, and that the affine set $W_p$ can be canonical chosen.
But if this is not the case, then you can consider the covering: $$\{W\mid\exists i,j\ \exists p\in U_i\cap V_j\text{ and }p\in W\subseteq U_i\cap V_j\text{ is affine and open}\}$$
Namely, follow the motto of choiceless math: When you don't know what to choose, take everything!
