The following is taken from Introduction to Topological Manifolds by John Lee
I am trying to verify that $\mathcal{T}$ is in fact a topology on $X^*$.
Note that definition I'm working with for the disjoint union is the following $X_1 \sqcup X_2 = \{ (x, \alpha) \ | \ x \in X_{\alpha} \ \text{ and } \ \alpha \in \{1, 2\}\}$. In the question below I set $X_1 = X$ and $X_2 =\{\infty\}$.
Also note that there is a hidden abuse of notation in the above quotation, which is the following. Letting $\phi_1 : X_1 \to X^*$ be the canonical embedding $\phi(x) = (x, 1)$ saying that
"$X^* \setminus U$ is a compact subset of $X$"
really means that $\phi^{-1}_1[X^* \setminus U]$ is a compact subset of $X$.
I'd just like to know if my reasoning below is correct. If we let $U$ be an open subset of $X$ and I pick $V \subseteq X^*$ such that $\phi^{-1}_1[X^* \setminus V]$ is a compact subset of $X$ then we have the following
$$U \cup V \not\subseteq X \ \ \ \ \text{and} \ \ \ \ U \cup V \not\subseteq X^*$$
because $U \cup V$ contains at least one point of $X^*$ and $U \cup V$ contains at least one point of $X$.
We need $U \cup V$ to be contained in $\mathcal{T}$. For this to happen $U \cup V$ must either be an open subset of $X$ or $\phi^{-1}_1[X^* \setminus (U \cup V)]$ needs to be a compact subset of $X$. But now $U \cup V$ certainly can't be a subset of $X$ so we have to have $\phi^{-1}_1[X^* \setminus (U \cup V)]$ to be a compact subset of $X$.
So to that end by elementary set theory we have $$\phi^{-1}_1[X^* \setminus (U \cup V)] = \phi^{-1}_1[(X^* \setminus U) \cap (X^* \setminus V)] = \phi^{-1}_1[X^* \setminus U] \cap \phi^{-1}_1[X^* \setminus V] \\= (X \setminus \phi^{-1}_1[U]) \cap \phi^{-1}_1[X^* \setminus V]$$
But now notice that in the expression above we have $\phi^{-1}_1 = \{x \in X \ | \ \phi(x) \in U\} = \emptyset$ so we have $(X \setminus \phi^{-1}_1[U]) \cap \phi^{-1}_1[X^* \setminus V] = X \cap \phi^{-1}_1[X^* \setminus V]$ which is the intersection of two closed sets in $X$ and is thus closed and hence compact in $X$ by Haursdoffness.
My question is if all of my arguments made above are correct. If they are not or if I am overcomplicating things please let me know. I am well aware that this does not verify that $\mathcal{T}$ is not a topology on $X^*$ nor does it show that arbitrary unions of open sets of $X^*$ are open in $X^*$. Finally please do not provide a proof that $\mathcal{T}$ is a topology on $X^*$ as I feel this is an important exercise and I want to complete the proof on my own.
EDIT: If $U$ is an open subset of $X$, is it then correct to define $\mathcal{T}$ as: $$\mathcal{T} = \{ \phi_1[U] \ | \ U \ \text{ is an open subset of } X\} \cup \{U \subseteq X^* \ | \ \phi_1^{-1}[X^* \setminus U] \text{ is a compact subset of } X\}$$
Is this what the author meant?
I think you're making too much of the "disjoint union" part. Just write union, and the only thing one should have is that $\infty \notin X$, it is a "new point". And given a set, it's always possible to find an element (really a set, in ZFC, where everything is a set) that is not a member of that given set.
So then we just can write given $(X,\mathcal{T})$, $X^\ast = X \cup \{\infty\}$ with topology $$\mathcal{T}^\ast = \mathcal{T} \cup \{U \subseteq X^\ast: X^\ast\setminus U \subseteq X \land X^\ast \setminus U \text{ compact in } X\}$$
where the $\subseteq X$ part in the latter part is to empahasize that $\infty \in U$ (or else $X^\ast \setminus U$ would contain $\infty$ and not be a subset of $X$). Also the Hausdorffness of $X$ ensures that the compact subsets of $X$ are closed, and so we're adding no new open subsets of $X$, i.e. $\mathcal{T}^\ast$ restricted to $X$ is just $\mathcal{T}$ again.
No need for disjoint unions and embeddings.