If it were true for C^1 curves, then since a Jordan curve is compact, it may be weierstrass approximated by a series of C^1 curves (indeed by curves whose component functions are polynomials) such that the series converges uniformly to the given jordan curve. Then by assumption, each curve in the sequence contains 4 points forming a square, and the sequence of squares can be regarded as (eventually) a sequence in the (sequentially) compact space of the 4-fold product of any closed epsilon enlargement of the area bounded by the original jordan curve. It follows that the sequence of squares contains a convergent subsequence, which can be shown to be a square lying on the original jordan curve.

Thus, proving the C^1 case proves the general case.

## no.

If it were true for C^1 curves, then since a Jordan curve is compact, it may be weierstrass approximated by a series of C^1 curves (indeed by curves whose component functions are polynomials) such that the series converges uniformly to the given jordan curve. Then by assumption, each curve in the sequence contains 4 points forming a square, and the sequence of squares can be regarded as (eventually) a sequence in the (sequentially) compact space of the 4-fold product of any closed epsilon enlargement of the area bounded by the original jordan curve. It follows that the sequence of squares contains a convergent subsequence, which can be shown to be a square lying on the original jordan curve.

Thus, proving the C^1 case proves the general case.