The theorem is not trivial. Everything hangs on the fact that the precise mathematical definition of "continuous" is extraordinarily broad; it allows for curves far stranger and more poorly-behaved than any typical "continuous" (in the intuitive sense) curve. Indeed it becomes considerably easier under any reasonable additional hypotheses, although as John Stillwell points out on MO one can force these hypotheses (e.g. piecewise-linear) to hold anyway. (I am somewhat surprised that he was able to avoid the full strength of the Jordan curve theorem, which is equally intuitive but really requires a lot of work to show for precisely the reasons I just gave.)
In general, there must necessarily exist short theorems with long proofs. More precisely, if $f(n)$ denotes the longest proof in a reasonably powerful formal system of a theorem of length $n$, then $f(n)$ grows faster than any computable function.
Proof. Suppose $f(n) \le g(n)$ where $g$ is computable. Then we can find an algorithm that finds proofs of theorems: simply test all proofs of length up to $g(n)$. But for a reasonably powerful formal system (one that can talk about Turing machines) such an algorithm cannot exist because of the unsolvability of the halting problem.
Whether this implies that there are intuitive results with non-intuitive proofs is up for debate. The examples I can think of (including the top-voted examples in the MO thread) are similar to yours; the result is clear intuitively because we have certain cases in mind, but the definitions are broad enough that some work is necessary to cover all cases.