How can the set of all graphs be the same size as its power set? Supose a set $G$ containing all graphs, either finite or infinite. If I take a set $S\in P(G)$ and draw all the graphs in $S$ on a paper I can always just combine them and create one posibly disconnected graph which is an element of $G$. So the combination of some graphs is itself a graph, meaning that $G=P(G)$.
How can this be posible or where is the error?
 A: You're not mistaken. You just proved that there is no "set of all graphs". In fact, there is no set of all singletons either. And so definitely there will be no set of all graphs.
In some set theories which admit a universal set (e.g. Quine's New Foundations), this is still not a problem, because the function mapping a set on its power set is not itself a set, so there is no contradiction to Cantor's theorem. (The reason this is happening is that the existence of a universal set means that not every formula can be used to define a subset.)
(Also, one minor mistake you have here, is that you didn't actually prove that $G=\mathcal P(G)$, but rather that there is a surjection from $G$ onto its power set. Note that a graph is not just a set, it's an ordered pair of set [of vertices] and a set of edges. So it isn't usually the case that a set of ordered pairs is itself an ordered pair.)
A: There are (at least) two issues:


*

*When you combine the graphs, graphs usually only have finitely many vertices and edges, while $S\in\mathcal{P}(G)$ may have infinitely many subgraphs.  So, does $G$ contain infinite graphs?  (see @HenningMakholm's comments below why $G$ can't contain infinite graphs.)

*Suppose that graph $A\in G$ is a disjoint union of two graphs $B$ and $C$.  Then using your construction, there are two ways to get $A$, $\{A\}\in\mathcal{P}(G)$ and $\{B,C\}\in\mathcal{P}(G)$.  Therefore, your map is not injective.  Hence, the argument does not show that $G=\mathcal{P}(G)$, only that there is a surjective map onto $G$.
