Suppose that your drawing DOES have a path that does what you say.
Any vertex is either a start or end vertex or an interior (visited along the way) vertex. If it's an interior vertex, it has an EVEN number of edges meeting it, because for each visit, there's the edge you arrived on and the edge you left on.
For a start/end vertex, it has an ODD number of edges meeting it. [But see below].
Conclusion: for a drawing with a suitable path, there are exactly two vertices with an odd number of edges meeting them. But in your example, there are 4 such vertices. Hence, no suitable path.
Revised and improved answer:
Look at each vertex of the drawing, and suppose that you had a path that traverses each edge exactly once. Begin by drawing a circle around the start and end points of the path (the two circles may be at the same vertex). At every non-circled vertex, the path arrives and leaves the same number of times, so all non-circled vertices have even degree. Conclusion: In the whole drawing, there are at most two vertices of odd degree. Your square with diagonals has 4. So the supposition that there is a "nice" path in this drawing must be incorrect.