One point compactification is compact

Question

I am starting to study the one point compactification of a $T_1$ space $S$, namely $S^*=S \cup\{\infty\}$.

I understand most of it except the part about $S^*$ being compact.

1. Given $C=\{A/A$ is open on $S^*$ $\}$ an open cover for $S^*$, at least one lets call it $A_{1}$ contains $\infty$.

2. Take all $A\in C$ such that $A \neq A_1$ then since $S^*-A_1=B$ is closed and compact on $S$ any open cover for $B$ has a finite subcover.

3. Also take $D=\{A\cap S/ A\in C\}$ then $D$ covers $B$. By the previous step $B=\cup(A_i\cap S)$ for $i=2,...,n$.

4. Now $\{A_1,A_2,...,A_n\}$ covers $S^*$.

I need help understanding 3 and 4. Thanks in advance.

Answer 1

There is a typo in 3). Instead of $A\cup S$ it should be $A\cap S$. Since $B \subset S$ and $B \subset S^{*}$ 3) follows from the fact that $D$ covers $S^{*}$. 4) follows from the fact that $S^{*} \subset B \cup A_1$.