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.
Here is my Math Stack Exchange question on the argument given in the preceding paragraph.
Second, suppose that $U$ contains $p$. Since $C = Y – U$ is closed in $Y$, it is compact as a subspace of $Y$. Because $C$ is contained in $X$, it is a compact subspace of $X$. Then because $X$ is a subspace of $Y^\prime$, the space $C$ is also a compact subspace of $Y^\prime$. …
Now my question is as follows:
Although $C$ is compact as a subspace of the topological space $X$ when $X$ has the subspace topology that $X$ inherits from $Y$, how do we gaurantee from this that $C$ would also be compact as a subspace of $X$ but when $X$ has the subspace topology that $X$ inherits from $Y^\prime$?
Best Answer
$X$ has a fixed topology. If you take any bigger space $Z \supset X$ such that $X$ is a subspace of $Z$, then by definiton the subspace topology on $X$ inherited from $Z$ is the original topology on $X$. The situation would be different if we only require that the underlying set of $X$ is a subset of $Z$. In that case the subspace topology on $X$ inherited from $Z$ can be anything; there is no relation to the original topology on $X$.