Suppose on the contrary that each unit square has at least two red edges. Each red edge is part of at most two unit squares. Therefore
$$2n^2 \le \sum_{\text{unit squares}} \text{(red edges of the square)} = \sum_{\text{red edges}} \text{(unit squares containing the edge)} \le 2n^2.$$
This implies that each unit square has exactly two red edges and that there are a total of $n^2$ red edges.
We colour each of the unit squares black and white such that no two unit squares which share a common edge have the same colour (like in a chess board). Note that each red edge is part of exactly one black square and one white square. Since every unit square has exactly two red edges, the number of white squares is therefore $n^2/2$, a contradiction since this is not an integer.
This shows that there is a unit square with at most one red edge.