Here is Prob. 5, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:
If $f \colon X_1 \rightarrow X_2$ is a homeomorphism of locally compact Hausdorff spaces, show $f$ extends to a homeomorphism of their one-point compactifications.
My Attempt:
Here is Theorem 29.1 in Munkres.
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 is a homeomorphism of $Y$ with $Y^\prime$ that equals the identity map on $X$.
The proof of Theorem 29.1 shows that the collection of the open sets of $Y$ consists of (1) all the open sets $U$ of $X$, and (2) all sets of the form $Y - C$ where $C$ is a compact subspace of $X$.
Following the proof of this theorem, Munkres states:
If $X$ should happen to be compact, then the space $Y$ of the preceding theorem is not very interesting, for it is obtained from $X$ by adjoining a single isolated point. However, if $X$ is not compact, then the point of $Y - X$ is a limit point of $X$, so that $\overline{X} = Y$.
After that, Munkres gives this definition.
If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y - X$ equals a single point, then $Y$ is called the one-point compactification of $X$.
Finally, Munkres states the following.
We have shown that $X$ has a one-point compactification $Y$ if and only if $X$ is a locally compact Hausdorff space that is not itself compact. We speak of $Y$ as "the" one-point compactification because $Y$ is uniquely determined up to a homeomorphism.
Now here is my attempt at Prob. 5, Sec. 29, in Munkres.
As $f \colon X_1 \rightarrow X_2$ is a homeomorphism of topological space $X_1$ with topological space $X_2$, so
(1) the map $f$ is a bijective mapping of $X_1$ onto $X_2$,
(2) the image set $f\left(U_1 \right)$ is an open set in $X_2$ whenever $U_1 \subset X_1$ and $U_1$ is an open set in $X_1$, and
(3) the inverse image $f^{-1}\left( U_2 \right)$ is an open set in $X_1$ whenever $U_2 \subset X_2$ and $U_2$ is an open set in $X_2$.
Let $Y_1$ and $Y_2$ be the one-point compactifications, respectively, of $X_1$ and $X_2$.
For each $r = 1, 2$, the open sets of $Y_r$ are as follows: (1) all sets $U_r$ that are open in $X_r$, and (2) all sets of the form $Y_r - C_r$ where $C_r$ is a compact subspace of $X_r$.
For each $r = 1, 2$, let $p_r$ be the single point of $Y_r - X_r$.
Now let $\tilde{f} \colon Y_1 \rightarrow Y_2$ be the mapping defined as $$ \tilde{f}(y) \colon= \begin{cases} f(y) \ & \mbox{ if } \ y \in X, \\ p_2 \ & \mbox{ if } \ y = p_1. \end{cases} \tag{Definition A} $$
We note that, as the mapping $f \colon X_1 \rightarrow X_2$ is bijective, so also is the mapping $\tilde{f} \colon Y_1 \rightarrow Y_2$.
Now let $V_1$ be any open set in $Y_1$. We show that $\tilde{f}\left( V_1 \right)$ is an open set in $Y_2$. The following two cases arise.
Case 1. If $V_1$ is an open set in $X_1$, then we find that $$ \tilde{f}\left( V_1 \right) = f\left( V_1 \right), $$ which is an open set in $X_2$ and hence this set is also open in $Y_2$.
Case 2. If $V_1 = Y_1 - C_1$ for some compact subspace $C_1$ of $X_1$, then we have $$ \tilde{f} \left( V_1 \right) = \tilde{f} \left( Y_1 - C_1 \right) = \tilde{f} \left( Y_1 \right) - \tilde{f} \left( C_1 \right) = Y_2 - f \left( C_1 \right), \tag{1} $$ because $\tilde{f}$ is bijective and also because $C_1 \subset X_1$.
Now as $C_1$ is a compact subspace of $X_1$ and the map $f \colon X_1 \rightarrow X_2$ is continuous, so the image set $f \left( C_1 \right)$ is a compact subspace of $X_2$. Therefore $Y_2 - f \left( C_1 \right)$ is an open set of $Y_2$. That is, $\tilde{f} \left( V_1 \right)$ is open in $Y_2$, by (1) above.
Alternatively, as $f \left( C_1 \right)$ is a compact subspace of $X_2$ and as $X_2$ is a subspace of $Y_2$, so $f \left( C_1 \right)$ is a compact subspace of $Y_2$ also.
And, as $Y_2$ is a Hausdorff space and as $f \left( C_1 \right)$ is a compact subspace of $Y_2$, so $f \left( C_1 \right)$ is closed in $Y_2$. Therefore the set $Y_2 - f \left( C_1 \right)$ is open in $Y_2$; that is, $\tilde{f} \left( V_1 \right)$ is open in $Y_2$. [Refer to (1) above.]
Thus we have shown that, for any open set $V_1$ in $Y_1$, the image set $\tilde{f}\left( V_1 \right)$ is also open in $Y_2$. Therefore $\tilde{f}$ is an open map.
Using a symmetric argument to show that the inverse mapping $\tilde{g} \colon Y_2 \rightarrow Y_1$ of $\tilde{f}$ is also open, we can conclude that the map $\tilde{f}$ is continuous.
Hence $\tilde{f} \colon Y_1 \rightarrow Y_2$ is a bijective continuous mapping that is also open. Hence $\tilde{f}$ is a homeomorphism.
Finally, from (Definition A) it is evident that the homeomorphism $\tilde{f} \colon Y_1 \rightarrow Y_2$ is an extension of the homeomorphism $f \colon X_1 \rightarrow X_2$.
Is this proof correct? If so, is it rigorous enough? Or, is there something lacking in my reasoning? Is there any detail that I have omitted?