I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.
For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of I. Rivin, Combinatorial Optimization in Geometry Adv. in Appl. Math. 31 (2003), no. 1, 242–271. (There is a free arxiv.org version, in case that matters.)
