Let $X$ a topological space and $\infty \in X$ with the topology given by $\tau=\lbrace A \subset X \mid \infty \not \in A, X \setminus A \text{ finite } \rbrace $ Interpret $X$ as a one point compactification.
Let $M=X \setminus \lbrace \infty \rbrace$ and $m\in M$. Equipped with the topology $\tau=\lbrace U \subset M \mid X \setminus U \, \, \text{Is finite}, m\not \in U\rbrace $.
I claim that $M$ is locally compact and Hausdorff.
In case of can prove it I like see that $\overline{M}=X$ and hence $X$ was the compactification of $M$.
I have a hour trying to prove that $M$ is locally compact, but I don´t get it. Otherwise is easy see that $M$ is Hausdorff.
Any advice of how to construct $M$ or what things I forget consider for makes $X$ the compactification of $M$.
The definition of $\tau$ is very poorly written. I would normally interpret it to mean that $\tau$ is the family of all subsets of $X$ that have finite complements and do not contain $\infty$, but that is not a topology on $X$, since it does not include the set $X$. Apparently the comma is being (mis)used to stand for or, and what is actually intended is that
$$\tau=\{A\subseteq X:\infty\notin A\text{ or }X\setminus A\text{ is finite}\}\,.$$
Now let $M=X\setminus\{\infty\}$. Then every $A\subseteq M$ satisfies the condition $\infty\notin A$, so every subset of $M$ is open in $X$. Now you should have no trouble showing that $M$ with the relative topology that it inherits from $X$ is locally compact and Hausdorff.
Your $\{U\subseteq M:X\setminus U\text{ is finite and }m\notin U\}$ is not a topology, since it does not include $M$, and is not the set of $A\cap M$ such that $A\in\tau$ no matter how the comma in the original definition of $\tau$ is interpreted.