Is there a Dihedral group of order 4?

You could interpret $D_4$ as symmetries of a $2$-gon, which has 2 vertices connected by 2 edges. Then swapping the edges is one generator (which has order 2) and swapping the vertices is another generator (which has order 2), so $D_4 \cong Z/2 \times Z/2$ in this case.

As $$D_{2n}=\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ so $$D_4=\langle x,y\mid x^2=y^2=(xy)^2=1\rangle$$ so $$D_4/\langle x\rangle\cong\mathbb Z_2=\langle y\rangle$$ But $\langle y\rangle$ is normal in $D_4$ so $D_4\cong\mathbb Z_2\times\mathbb Z_2$

Sure. In that case, $D_{4}$ is the group of symmetries of a $2$-gon, and is in fact isomorphic to the Klein 4-group, $V_{4}$. However, unless I've misunderstood you, I think you ought to think about the answer to your question (the one you linked to) again - it looks perfectly good to me.