Chromatic number of the power set Let $X$ be a non-empty set. Consider $\mathcal{P}(X)$, the power-set of $X$. We say that $a,b \in \mathcal{P}(X)$ form an edge if and only if their symmetric difference is a singleton, i.e.  $\textrm{card}((a\setminus b) \cup (b\setminus a)) = 1$.
It is clear that for a finite set $X$ the resulting graph has chromatic number 2: color those subsets of $X$ with an even number of members with color 1, and the rest with color 2.
For $X$ infinite, does the resulting graph still have chromatic number 2?
 A: Yes, for $X$ infinite the resulting graph also has chromatic number 2.  To see this, just use the fact that a graph is bipartite if and only if it does not contain an odd cycle (this remains true for infinite graphs).  An odd cycle in your graph quickly leads to a contradiction due to parity reasons.  
A: The affirmative answers by Tony Huynh and Gjergji Zaimi are correct, but it may be worth noting that both depend on the axiom of choice.  For example, although each of the components (as described in Gjergji's answer) is 2-colorable, it has (exactly) two 2-colorings, and you need to choose one, in each component, in order to 2-color the whole graph.  In some models of set theory without choice, for example Solovay's model where all sets of reals are Lebesgue measurable, there is no way to make these choices and your graph (even for $X=\mathbb N$) is not 2-colorable.  Even in the presence of the axiom of choice, so that your graph is 2-colorable, there may not be a definable 2-coloring, even allowing arbitrary ordinal numbers and arbitrary real numbers as parameters in the definition.
A: Your graph is going to be disconnected. Every connected component is a graph with vertices having finite symmetric difference with respect to some base set. Once you observe this, you can still use the parity argument, and split the sets according to whether they differ from the base set on an even or odd number of places. 
