This problem came up on my proofs class, but there appears to be some extra condition that isn't specified, because I came up with a counter example. Note that it has 3 outside doors and the middle room has 2 doors. 
My professor couldn't help, since this question was written by another prof when he was teaching the class. I understand the handshaking lemma, which I think might be relevant but I don't really see how. Can someone tell me what element of graph theory this problem is getting at and why my counterexample is wrong?
I think the phrasing of the title is bad. The generalization of the the question is: a house with an odd number of exterior doors must have at least one room with an odd number of doors. In other words, it's not possible to have a house where each room has an even number of doors.
To see this, the outside of the house can be unified into one vertex, and all rooms get their own vertex. Each door is represented by an edge. Now sum over the degree of each vertex. The outside vertex has odd degree $d_o$, whereas each room has degree $d_r$. So $d_o+\sum_{r=1}^Nd_r=2E$.
Then $2E-d_o$ is odd. This implies you must have at least one room with an odd number of doors.
So your counterexample doesn't work as you have a room with 1 door.