Here is Theorem 29.1 in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be a [topological] space. Then $X$ is locally compact Hausdorff if and only if there exists a [topological] space $Y$ satisfying the following conditions:
(1) $X$ is a subspace of $Y$.
(2) The set $Y – X$ consists of a single point.
(3) $Y$ is a compact Hausdorff space.
If $Y$ and $Y^\prime$ are two spaces satisfying these conditions, then there exists a homeomorphism of $Y$ with $Y^\prime$ that equals the identity map on $X$.
Here is the initial part of Munkres' proof:
Step 1
We first verify uniqueness. Let $Y$ and $Y^\prime$ be two [topological] spaces satisfying these conditions. Define $h : Y \longrightarrow Y^\prime$ by letting $h$ map the single point $p$ of $Y – X$ to the point $q$ of $Y^\prime – X$, and letting $h$ equal the identity on $X$. We show that if $U$ is open in $Y$, then $h(U)$ is open in $Y^\prime$. Symmetry then implies that $h$ is a homeomorphism.
First, consider the case where $U$ does not contain $p$. Then $h(U) = U$. Since $U$ is open in $Y$ and is contained in $X$, it is open in $X$. Because $X$ is open in $Y^\prime$, the set $U$ is also open in $Y^\prime$, as desired.
Now my question is as follows:
Since $U$ is open in $Y$ and contained in $X$, it is open in $X$ — when $X$ is regarded as a subspace of $Y$ and not necessarily when $X$ is regarded as a subspace of $Y^\prime$. Because $X$ is open in $Y^\prime$, how does it follow from this that $U$, which is open in the subspace $X$ of $Y$ and not necessarily open in the subspace $X$ of $Y^\prime$, is also open in $Y^\prime$? After all, until the final conclusion of the theorem has actually been established, $Y$ and $Y^\prime$ are different (i.e. potentially non-homeomorphic) topological spaces.
Best Answer
The subspace topologies that $X$ inherits from $Y$ and from $Y'$ are identical, so it's rather trivial that if $U\subseteq X$ is an open subset in one of the topologies, it's open in the other.
As for why this implies openness in $Y'$: In general, given that $X\subseteq Y'$ is open and $U\subseteq X$, the following two are equivalent:
The upward implication is a straight-forward application of the definition of subspace topology, and doesn't use the openness of $X$.
The downward implication may be proven as follows: By the definition of subspace topology, there is an open subset $U'\subseteq Y$ such that $X\cap U'=U$. But the left-hand side is an intersection of two open subsets of $Y'$. So it's open in $Y'$.