Image of a patch - an open set?

O'Neill's Elementary Differential Geometry, exercise 4.3.13 (first edition) poses the following problem:

"Prove that if $y:E\rightarrow M$ is a proper patch, then y carries open sets to open sets in M. Deduce that if $x D\rightarrow M$ is an arbitrary patch, then the image x(D) is an open set in M. (Hint: To prove the latter assertion, use corollary 3.3)"

Corollary 3.3 is "If x and y are patches in a surface M whose images overlap, then the composite functions $x^{-1}y$ and $y^{-1}x$ are differentiable mappings defined on open sets of $E^2$.

I am doubtful of the first part of the problem. For the second part, I conclude that the intersection of the images of an arbitrary patch x and a proper patch y is an open set. But how does that imply that the image of x is an open set?