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Cut the octagon for $S_3$ along a diagonal joining the appropriate two vertex classes. The two resulting polygons can be reattached along that diagonal in the opposite order. Reading the new boundary word gives the square wordwhich is exactly the edge identification displayed for $S_4$. The cut-and-paste map is affine on the two pieces and respects every paired edge, so it descends to a homeomorphism $S_3\cong S_4$. Both are the Klein bottle.

After deleting an open disc, the punctured Klein bottle can be realized as a boundary connected sum of two embedded Möbius strips, and hence embeds in ${\mathbb{R}}^3$. The closed Klein bottle cannot embed: every connected closed surface embedded in ${\mathbb{R}}^3$ is two-sided and therefore orientable, by the Jordan-Brouwer separation theorem, whereas the Klein bottle is nonorientable. Thus