Well known result in general topology is that locally compact Hausdorff space $X$ can be embedded in compact Hausdorff space $Y$ such that $Y = X\cup\{\infty\}$. And this is unique up to homomorphism. Standard proof goes that if you take two of those spaces, say $Y, Y'$, you can define a bijection with identity on $X$ and mapping additional one point to each other. So, by symmetry, it suffices to show that if $U$ open, $h(U)$ is open.
I have a question if $\infty\notin U$. So the argument goes like this (quoting from Munkres, Topology 3rd ed, Thm 29.1): $h(U) = U$. Since $U$ is open in $Y$ and is contained in $X$, it is open in $X$. Because $X$ is open in $Y'$ (we use Hausdorff space here), the set $U$ is also open in $Y'$ (???). Problem I have with this is that $U$ is open in $X$ in topology generated by $Y$. How do we know that it is open in $X$ in topology generated by $Y'$? Topology in $Y'$ may very well be $\{\emptyset, X, Y'\}$. Identity function is obviously not continuous if topologies are different.
Can someone please reveal the obvious thing that I'm missing here?
One thing to keep in mind: To say that $X$ is embedded in $Y$ means that the given topology on $X$ is identical to the subspace topology on $X$ relative to $Y$ (what you are called the topology on $X$ that is "generated by $Y$", but I am trying to keep closer to Munkres' language of the subspace topology). Similarly $X$ is embedded in $Y'$, so again the given topology on $X$ is identical to the subspace topology on $X$ relative to $Y'$.
Using your example where the topology on $Y'$ is $\{\emptyset, X, Y'\}$, then it follows that the given topology on $X$ is $\{\emptyset,X\}$, and from that it follows that the topology on $Y'$ is $\{\emptyset,X,Y'\}$.