Assuming that K is smooth or locally flat, there is a nice geometric reason based on transversality.
The Klein bottle is not orientable. If it embedded in R3 so as to separate a small epsilon neighborhood of its image, then the neighborhood would be homeomorphic to a trivial line bundle over the Klein bottle, K times (-1,1), and therefore the neighborhood would also be non-orientable. But since any 3-dimensional submanifold of R3 is orientable this cannnot happen.
So the Klein bottle embedding would have to be non separating in this neighborhood. But this means there is a curve C in the neighborhood that intersects the embedded Klein bottle once, transversally.
Now C bounds a disk D in R3 since R3 is simply connected. (Bounding any surface is enough). One can make D transverse to the Klein bottle K, leaving boundary D fixed. One then see that K intersects D in a union of arcs and closed curves, with a single boundary point. But any such collection has an even number of boundary points, a contradiction.
This argument works more generally to show a closed nonorientable surface cannot be embedded in a manifold 3-manifold M with H1(M;Z2) = 0.