I've recently read some papers and books involving simply connected domains in Euclidean space (dimension at least 2), where domain is an open connected set. The usual definition is a (connected) set for which every continuous closed curve is (freely) contractible while some authors only require that every continuous simple closed curve is contractible. The authors who define simple connectedness using simple closed curves do so in order to use Stokes' Theorem or the Jordan curve theorem somewhere in the sequel; however, they never mention (not even with a reference) that their definition is equivalent to the usual one! My question is if there is a proof written down somewhere (with all the details) proving the equivalence (for domains in $\mathbb{R}^n$ with $n \geq 2$)? If not, does someone know of an "easy" proof using a minimal amount of knowledge, say that of a first course in topology?

As pointed out by Pierre and Paul in comments, there are several standard ways to deal with this kind of issue. A good answer really depends what you're assuming you start from, and where you're trying to go to. The Jordan curve theorem and Stoke's theorem are both fairly sophisticated and difficult for beginners to grasp, so it's a bit hard to see how only analyzing embedded curves is streamlining anything, except perhaps helping with people's intuitive imagesbut even so, it may do more harm than good. Perhaps it's worth pointing out that this statement is false in greater generality, for instance for closed subsets of $\mathbb R^3$. Here's an example in $\mathbb R^3$: consider a sequence of ellipsoids that get increasingly getting long and thin; to be specific, they can have axes of length $2^{k}$, $ 2^{k}$ and $2^k$. Stack them in $\mathbb R^3$ with short axes contained in the $z$axis, so each one touches the next in a single point with long axes parallel to the $x$axis, and let $X$ be their union together with the $x$axis. Any simple closed curve in $X$ is contained in a single ellipsoid, since to go from one to the next it has to cross a single point, so every simple closed curve is contractible. Anyway, here are some lines of reasoning that can overcome whatever hurdle needs to be ovrcome:



The equivalence is conceptually easy: each closed curve is a union of simple closed curves. If you can contract each simple closed curve, you can contract the whole curve. Each simple closed curve also lives in the set of closed curves, so the equivalence the other direction is simple. This sort of proof shouldn't be too hard for you to construct, assuming you have the knowledge of a first course in topology. Some care might need to be taken in constructing the explicit homotopy and in dealing with a curve which has infinitely many selfintersection points, but these are both issues you should have seen in such a first course in topology. 

