As a student, I was taught that the Jordan curve theorem is a great example of an intuitively clear statement which has no simple proof.
What is the simplest known proof today?
Is there an intuitive reason why a very simple proof is not possible?
As a student, I was taught that the Jordan curve theorem is a great example of an intuitively clear statement which has no simple proof. What is the simplest known proof today? Is there an intuitive reason why a very simple proof is not possible? 


There's a short proof (less than three pages) that uses Brouwer's fixed point theorem, available here: The Jordan Curve Theorem via the Brouwer Fixed Point Theorem The goal of the proof is to take Moise's "intuitive" proof and make it simpler/shorter. Not sure whether you'd consider it "nice," though. 


It depends on what you mean by "simple". If you know homology, the proof is not very hard (less than 1 page), see for example, section 2.B ("Classical Applications") of Hatcher's book "Algebraic Topology". 


Carsten Thomassen's proof is relatively simple: Carsten Thomassen, The JordanSchönflies theorem and the classification of surfaces. Amer. Math. Monthly 99 (1992), no. 2, 116130. By the way, the Jordan Curve Theorem has a formal proof (one that can be checked by a computer): Thomas C. Hales, The Jordan curve theorem, formally and informally. Amer. Math. Monthly 114 (2007), no. 10, 882894. Hales bases the formal proof on Thomassen's. The following is a survey on the older papers on the subject: H. Guggenheimer, The Jordan curve theorem and an unpublished manuscript by Max Dehn. Archive for History of Exact Sciences 17 (1977), 193200. 


Several proofs are here: http://www.maths.ed.ac.uk/~aar/jordan/index.htm Among them, Tverberg's (1980) could (and should) be mentioned. But, after reading (and reading) http://www.math.sunysb.edu/~bishop/classes/math401.F09/HalesDefense.pdf , I really like Jordan's proof itself. 


There is a proof of the Jordan Curve Theorem in my book "Topology and groupoids" which also derives results on the PhragmenBrouwer Property. Also published as `Groupoids, the PhragmenBrouwer property and the Jordan curve theorem', J. Homotopy and Related Structures 1 (2006) 175183. The van Kampen Theorem for the fundamental groupoid on a set of base points is used to prove that if $X$ is pathconnected and the union of open path connected sets $U,V$ whose intersection has $n$ path components, then the fundamental group of $X$ contains the free group on $n1$ generators as a retract. May 30: The question asks why there is not a simple proof. Perhaps the following Figure 9.10 from the above book will explain why a proof is not expected to be so so easy; how do you decide whether a point in the middle is inside or outside? 


You should compare with: "Geometric Topology in Dimensions 2 and 3", Moise, Edwin E. (1977). SpringerVerlag and tell 


An elementary proof by means of nonstandard analysis (by reduction to the case of polygons) and elementary combinatorics is given in Kanovei & Reeken, A nonstandard proof of the Jordan curve theorem, RAE 1999, 24, 161170 


There's a remarkable elementary proof of the Jordan separation theorem, using only the fundamental group, due to Doyle. The proof is expounded in detail in Armstrong's book Basic Topology, Section 5.6. I think this approach could be extended to prove that there are two complementary components. If there were more, then by an application of Van Kampen's theorem, one could conclude that the fundamental group is a free group of rank $>1$, which would give a contradiction as in Doyle's argument. 


A nice and simple proof using mod 2 intersection theory is given in the book Differential Topology by Guillemin,Pollack. 

