I have the theorem below, I did some work but stuck at some point. Any help would be great.
Theorem. Let $U$ be an open subset of a compact Hausdorff space $X$ and $U^*$ is its one-point compactification. If $\phi:X \to U^*$ is defined by $\phi(x)=x$ if $x \in U$ and $\phi(x)= \infty$ if $x \in U^c$ then $\phi$ is continuous.
Proof. Let $V$ be open in $\tau_{U^*}$. Then if $\infty \notin V$ then we can say that $V \subset U$ and open in $\tau_{U}$. This implies that $V = U \cap W$ for some $W \in \tau_{X}$. So, $\phi^{-1}(V)=V$ which is open since both $U$ and $W$ is open in $X$.
If $\{\infty\} \notin V$, how can I show that $\phi^{-1}(V)$ is open in $X$?
If $\infty \in V$, then $U^* - V = U - V$ is compact. Since $X$ is Hausdorff, $U - V$ is closed in $U$. This implies that $U - (U - V)$ is open in $U$, thus open in $X$ because $U$ is open in $X$. Since $$\varphi^{-1}(V) = V - \{\infty\} = U - (U - V)$$ this is all we needed.