Let us consider the plane $P = \{(x, y, -1) \in \mathbb{R}^3\}$ in three dimensional space. It is a closed set, but because it isn't bounded, it isn't compact. Now take the sphere without a pole: $$S^2_* = \{(x, y, z) \in \mathbb{R}^3\mid x^2 + y^2 + z^2 = 1\} - \{(0,0,1)\}$$ It is bounded by the sphere itself, and it's also a closed set. Thus, it is a compact set. Then comes stereographic projection, giving a continuous bijection between $P$ and $S^2_*$ - but $S^2_*$ is compact and continuous functions take compact sets to compact sets! Clearly, $P$ is not compact, so something fishy is going on here.
My guess is that stereographic projection continuity is the problem - maybe I'm considering the wrong topologies on those sets. But ain't the subspace topology on $P$ the same as the metric topology on $\mathbb{R}^2$? I see open disks corresponding to the intersection of open balls with the plane. Or is the problem in the closeness of $S^2_*$? I guess the pole missing ain't an interior point of the complement..
The sphere without a pole $S^{2}_*$ is not a closed set because its complement is not open.
The complement is not open because it includes the sphere's pole, which violates the neighborhood condition for open sets.