There is one very simple proof for Jordan curve theorem http://www.maths.ed.ac.uk/~aar/jordan/maehara.pdf
But Maehara's proof contains one "trivial" step when we fit Jordan curve $J$ in a rectangular set $E=(-1,1;-2,2)$. Of course we can say that "let's choose our coordinates so that...", but that choise is equal to isomorphic transformation and when we are trying to prove a fundamental topological property, then one could argue that it's not at all clear that we can transform the plane when considering an arbitraty homeomorphism $h:S^1 \rightarrow \mathbb{R}^2$.
So is it trivial that an isomorphic transformation preserves the number of connected components?
An isomorphism is the topological category is usually called a homeomorphism, and there doesn't exist a homeomorphism $h:S^1 \to \mathbb{R}^2$ (for example, the former is compact and the latter is not); we could talk about something like a (topological) embedding, etc.
Anyway, you're asking whether homeomorphisms preserve connected components. That's true: A homeomorphism $f:X \to Y$ induces a bijection between the set of connected components of $X$ and the set of connected components of $Y$. (This is usually written in terms of the object $\pi_0$, but I'm going to work directly from definitions here.) A connected components of $X$ are those connected subspaces of $X$ that are maximal with respect to inclusion, and homeomorphisms preserve connectedness and respect inclusion. The latter is clear; for the former, note that a connected space is exactly one containing no proper subsets that are both closed and open, and homeomorphisms preserve all those conditions.