Set theoretic implications of constructions in Differential Geometry/ Topology In subjects like Differential Geometry/ General Topology one often constructs for each $x$ in a space $X$ a set $U_x$ satisfying certain properties. Examples where one does constructions like this:


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*Deducing that every open cover of a second countable space has a countable subcover.

*Proving that any covering space of a topological $n$ - manifold is second countable.


I am interested in whether there are any set theoretic issues like say use of the axiom of choice. In particular, how can we say "for each $x$ construct a set $U_x$.." simultaneously for all $x \in X$?
Also, my set theory is not so deep but I know that assuming that the set of all sets is a set leads to a contradiction in the style of Russell's paradox. However somehow it still makes sense to speak of a "maximal smooth atlas containing a given atlas" of a smooth manifold, or all closed sets containing a given set in a topological space. 
Why is there no problem in doing this and we don't get to a situation like Russell's paradox? 
 A: In general, one needs the axiom of choice to choose $U_x$'s as in your question, but in certain cases one can be more explicit and thus avoid the axiom of choice.  
Your second example requires some form of the axiom of choice.  Specifically, it is consistent with ZF to have a countable sequence of pairwise disjoint $2$-element sets $P_0,P_1,\dots$ with no selector (i.e., no set consisting of exactly one element from each of the pairs $P_n$).  In this situation, let $X$ be the union of the $P_n$'s, with the discrete topology, also give $\mathbb N$ the discrete topology, and let $f:X\to\mathbb N$ send the two elements of any $P_n$ to $n$. Then $f$ is a $2$-to-$1$ covering map, $\mathbb N$ is obviously second countable, but $X$ is not second countable (any basis for its topology would have to include all the singletons and would thus give an enumeration of $X$, contrary to the non-existence of a choice function for the pairs $P_n$).  
I suspect that your first example also requires some form of the axiom of choice, but I don't immediately see a counterexample.
EDIT: Now I see a counterexample for the first one.  Work in Cohen's first model for the failure of choice.  This model contains a set $A$ of mutually generic Cohen subsets of $\mathbb N$ such that $A$ has no countably infinite subset.  It follows from genericity that $A$ is a covering of $\mathbb N$ but any countable subfamily, being finite, fails to cover $\mathbb N$.  Giving $\mathbb N$ the discrete topology, we get an open cover $A$ with no countable subcover, even though $\mathbb N$ is trivially second countable.
