A continuous function that maps closed unit square to the unit open square

Solution 1:

The compactness argument is quite valid, $(0,1)^2$ is not closed so not compact and the continuous image of a compact space is compact.

The completeness argument is not correct. The continuous image of a complete metric space (even within the space) need not be complete, e.g. $\arctan(x): \Bbb R \to \Bbb R$ has image $(-\frac{\pi}{2}, \frac{\pi}{2})$. Completeness is a property of the metric, not of the topology and continuous functions need not preserve metric values (isometries do).

There are also more advanced topological properties that the closed square has and the open one does not, but I think compactness is the simplest solution to this problem.