Monologue

I'm thinking about a weird construction in topology: the ordered square. Think of it as a square on the ordinary euclidean plane with corners on the points <0, 0> and <1, 1>, boundaries included. We can define an order on the square by comparing two points p, q on that square the following way:

If two points p = <a, b> and q = <c, d> are on the same vertical, i.e., a = c, then we say that p < q if b < d as ordinary real numbers. If the points are not on the same vertical, perhaps we have a < c, then p < q. (Note that even though I'm using the same symbol '<' for order on the square and on the ordinary real numbers, they are different concepts altogether.)
This square as a topological space has the property of
connectedness (i.e., informally put, it's "all in one piece") but it doesn't have the property of
path-connectedness (i.e., any two points can be connected by a path).

I've been trying to think of an informal explanation for why it's not path-connected. It's easy to see (I think?) that the ordered square is path-connected locally, along every vertical including the boundary points since it has the same order type as the closed interval along verticals. But why is it not true globally? It seems as though any two points on distinct verticals are not path-connected because the closed interval is maybe "too small" to map between the two points. I'm stretching beyond what I actually know here but I think it's because while we may be able to order-embed(?) a closed interval into the ordered square easily along each vertical, we can't do that with endpoints on different verticals, perhaps because between any two verticals, there are uncountably many such verticals. Not even sure if any of these thoughts make sense.