As far as i know the simplest (in my opinion) self contained proof of the planar separator theorem is unpublished; it is a combination of the following (somewhat non-trivial) theorem with the "Baker layering" trick.
The treewidth of a planar graph of diameter d is <= 3d-1
[H. L. Bodlaender: A partial k-arboretum of graphs with bounded treewidth.
Theor. Comput. Sci., 209(1-2):1–45, 1998, Theorem 83]
Now to the "Baker layering" trick: do a BFS from an arbitrary vertex and number the layers
1,2....sqrt(n), 1, 2, ... sqrt(n) .. etc
Now there exists an i such that the total number of vertices living in layers labelled i is at most sqrt(n). Let the set of these vertices be S. Consider any connected component of G-S. Such a component must live in at most sqrt(n) consecutive layers, and thus it has diameter <= 2sqrt(n). Since planar graphs of diameter sqrt(n) have treewidth at most 3sqrt(n) we have that G-S has treewidth <= 6sqrt(n) and so G has treewidth <= 7sqrt(n). Tada.
There is a small silly bug in the proof above; the fact that a connected component of G\S lives inside sqrt(n) layers does not mean that the diameter is actually at most sqrt(n) ... but thus is easily fixed by contracting the layers up until the layer of this component into a single vertex. Then diameter of (a supergraph of the component) is at most 2sqrt(n), and we are done.