Proving that $(Y, \tau_\infty)$ is compact I have questions regarding the proof that I made about the following statement: "Let $(X,\tau_{X})$ be a topological space and $\lbrace \infty\rbrace$ an object that doesn't belong to X. Define $Y=X\cup\lbrace\infty\rbrace$ and
\begin{equation}
\tau_{\infty}=\lbrace U\subset Y|U\in \tau_X\text{ or }Y-U\text{ is compact and closed in } X\rbrace
\end{equation}
which is a topology on $Y$. Show that $(Y,\tau_{\infty})$ is compact.
Ok, here is my attempt of the proof: Let $C=\lbrace C_\alpha \rbrace_{\alpha \in \Lambda}$ be an arbitrary open cover of $Y$. Since $C_\alpha \in \tau_\infty$ for every $\alpha$, then $Y-C_\alpha$ is compact and closed in $X$ for every $\alpha$. Since $Y-C_\alpha$ is compact, let $C_x=\lbrace C_{\alpha_x}\rbrace_{\alpha_x\in \Lambda_x}$ be any open cover of $Y-C_\alpha$. Then there is a finite subcover $C_x^{\prime}=\lbrace C_{\alpha_{x},i}\rbrace_{i=1}^n$, with
\begin{equation*}
Y-C_\alpha \subset \cup_{i=1}^nC_{\alpha_{x},i}.
\end{equation*}
Now, select $\alpha_\infty$ such that $C_{\alpha_\infty}$ in $C$ contains the object $\lbrace \infty \rbrace$. Then,
\begin{equation}
Y-C_{\alpha_\infty}\subset \cup_{i=1}^nC_{\alpha_{x},i}
\end{equation}
Then clearly $\lbrace C_{\alpha_{x},i}\rbrace_{i=1}^n \cup \lbrace C_{\alpha_\infty}\rbrace$ is a finite subcover of $Y$. Hence, $(Y,\tau_\infty)$ is compact.
I just want to know if my proof is correct. I would appreciate any corrections or other suggestions in case if it is incorrect. Thank you!
 A: Your idea is correct, but you do not properly elaborate it. You have to find a finite subcover of $C=\lbrace C_\alpha \rbrace_{\alpha \in \Lambda}$.
You consider any open cover $C_x=\lbrace C_{\alpha_x}\rbrace_{\alpha_x\in \Lambda_x}$ of $Y-C_\alpha$ - but this has nothing to do with the given cover $C$.
So proceed as follows:

*

*Select $\alpha_\infty$ such that $C_{\alpha_\infty}$ in $C$ contains the object $\lbrace \infty \rbrace$.


*The set $Y-C_{\alpha_\infty}$ is compact and the collection $C'  = \{ C_{\alpha} \cap (Y-C_{\alpha_\infty}) \}$ is an open cover of $Y-C_{\alpha_\infty}$. Hence it has a finite subcover  $C_{\alpha_i} \cap (Y-C_{\alpha_\infty})$, $i = 1,\dots, n$. But then $\{ C_{\alpha_i} \mid i= 1,\dots,n,\infty\}$ is a finite subcover of $C$.
The space $Y$ is known as the Alexandroff compactification of $X$. See https://math.stackexchange.com/search?q=Alexandroff+compactification and https://en.wikipedia.org/wiki/Alexandroff_extension.
A: 
Since $C_\alpha \in \tau_\infty$ for every $α$, then $Y−C_\alpha$ is compact and closed in $X$ for every $\alpha$.

This is not true : if $C_\alpha \in \tau_X$, you don't yet know that the complement is compact.
Fortunately, since $\infty \in C_{\alpha_\infty}$, you have $C_{\alpha_\infty} \notin \tau_X$ and therefore its complement is closed and compact in $X$. The rest of your proof works from this point on.
