I like Jonah Ostroff short and sweet proof, but the key to it lies in the fact that there is not a bijection between the set $S_1$ of connected graphs and the set $S_2$ of disconnected graphs over $n$ labeled vertices for $n \ge 4$, as follows:
the complement of each disconnected graph is a connected graph (which Ostroff points out)
the complement of a connected graph can also be a connected graph
thus the cardinality of the set of connected graphs must be larger than the cardinality of the disconnected graphs, because while there is a one-to-one mapping of each disconnected graph onto a connected graph, there exist connected graphs which do not map to a disconnected graph
For example, for $n=4$:
Take the $12$ possible unidrected un-drected Hamiltonian paths of length $4$ on a graph over four labeled vertices.
The complement of each of these paths is also a hamiltonian path.
Since we know that the complement of a disconnected graph is obviously connected for $n>3$, then the number of connected graphs is at least equal to the number of disconnected graphs. Hoewever, since for $n>3$, the complements of at least some of the connected graphs are also connected graphs, that means that there must be more connected graphs than there are unconnected graphs.
The $12$ Hamiltonian paths are those connected graphs over $4$ vertices whose complements are also connect: thus the remaining $2^6 - 12 = 56$ 52$ graphs are divided into pairs of complement graphs which are connected and disconnected,
yielding a total of $26$ disconnected graphs, and $26+12=38$ connected graphs over the set of $64$ labeled graphs over $4$ labeled vertices.
The path graphs of length $n$ on the set of $n$ vertices are the canonical example of connected graphs whose complements are also connected graphs (for $n>3$).