Prove that the dual graph of any (planar) graph is connected I'd like to know if there's a standard proof that the dual graph of any planar graph is connected (or, if there's a counterexample, I'd like to know that too).  I've thought of a proof that might work (see below), but this seems, at the very least, like overkill.  Any suggestions to improve my proof, or any simpler or more transparent proofs for that matter, would be very much appreciated.
Also, if I flub any of the terminology, do let me know.

The proof I had in mind:
Suppose we have a (compact) embedding of $G$ into $\mathbb{R}^2$.  It suffices to show that there is a path (passing through edges) that connects any internal face to the external face.
With that in mind, consider an arbitrary internal face $F$, and a point $x_F \in F$.  We note that there are infinitely (uncountably) many lines passing through $x_F$.  However, for each vertex $v$ of $G$, there is a unique line that passes through $x_F$ and $v$.  Because there are more lines through $x_F$ than vertices, it follows that there exists a line through $x_F$ that does not pass through a vertex of $G$.
That is, there exists a line $x_F$ that passes only through the edges of $G$.  Since the graph is bounded and the line is unbounded, there exists a point $x_E$ of the line which is on the external face of the graph.  
If we follow the line from $x_F$ to $x_E$, we "describe a path" in the dual graph from $F$ to the external face.  
Thus, each vertex of the dual graph is connected to the vertex corresponding to the external face, which means that the dual graph must be connected.
 A: Maybe a nicer way is to use cycle space-cut space duality.
First proof: If $G$ is a graph with $n$ vertices, $m$ edges and $c$ components, then the dimension of the cycle space $\mathcal{C}(G)$ is $m-n+c$ and the dimension of the cut space $\mathcal{B}(G)$ is $n-c$. On the other hand, if $G$ is a planar graph with dual $G^*$, then $\mathcal{C}(G^*)=(\mathcal{B}(G))^*$.
So if $G$ is a connected planar graph, then $\dim(\mathcal{C}(G^*))=n-1=m^*-n^*+1$, where $n^*=2+m-n$ and $m^*=m$ are the number of vertices and edges of $G^*$. Hence $G^*$ is connected.
Second proof: Here is a more concrete way, with notation as above. The cycle space-cut space duality basically says that the edges of a circuit in $G$ (respectively $G^*$) correspond to the edges in a cut set in $G^*$ (resp. $G$). 
Now, let $T$ be a spanning tree of $G$ and consider the set $F\subseteq E(G^*)$ of "dual" edges corresponding to $E(G)\setminus T$. Of course, $F$ has $m-n+1=n^*-1$ edges. Also, if $F$ contains a cycle, then the edges of the cycle correspond to a cut set in $G$ not containing an edge of $T$, an impossibility. Hence $F$ is a spanning tree of $G^*$. In particular, $G^*$ is connected. 
