In this question cubes are topological spaces of the form $[0,1]^J$ with product topology and $[0,1]$ with usual topology. Further a space is a Tychonoff space if and only if it is a completely regular space. Also note that cubes are compact Hausdorff spaces.
In the wide look that two spaces that are homeomorphic are actually the same I was eager to classify certain spaces as subspaces of a cube, and made the following observations:
Tychonoff spaces are exactly the subspaces of cubes.
Compact Hausdorff spaces are exactly the subspaces of cubes that have a closed subset of the cube as underlying set.
Locally compact Hausdorff spaces are exactly the subspaces of cubes that have the intersection of a closed and an open subset of the cube as underlying set.
The first observation is a corollary of the Imbedding theorem for Tychonoff spaces (Munkres Th. 34.2).
The second observation follows from the fact that compact Hausdorff spaces are Tychonoff spaces. They are compact in the cube which is Hausdorff, so their underlying set is a closed set.
The third observation is important in my question. It is based on the statement that a space is homeomorphic to an open subspace of a compact Hausdorff space if and only if $X$ is locally compact Hausdorff (Munkres Cor. 29.4).
First question (a check of someone else never harms):
Are these observations correct?
Second question (the main):
Can locally compact Hausdorff spaces also be classified (up to homeomorphisms) as exactly the subspaces of cubes that have an open subset of the cube as underlying set?
So asked is actually whether I can replace my third observation (intersection of open and closed set) with this (more comfortable) one.
If the answer is "no" then of course I would like to see a counterexample.
Thank you for reading this already.