Let $X$ be a locally compact $($not compact$)$ Hausdorff space. Let $Y=X\cup \{0\}$ be the one point compactification of $X$. Now define a set $C_0(X)$ by:
$$C_0(X):=\left\{f \in C(X): f \text{ vanishes at infinity}\right\}.$$
Now we say that $f\in C(X)$ vanishes at infinity if for each $\varepsilon>0$ the set $\{\omega \in X: |f(\omega)|\geq \varepsilon\}$ is compact. Now consider $C(Y)$, the set of all continuous functions from $Y$ to $\mathbb C$. Now there is a statement in the middle of a proof that, a function in $C(Y)$ belongs to $C_0(X)$ if and only if $f(0)=0.$ I am not able to prove this statement.
My approach: $~''\Leftarrow''~$ Let $f \in C(Y)$ and $f(0)=0$. Now$f$ is continuous at $0$. Then for each $\varepsilon>0$ there exists a neighborhood $U_0$ of $0$ such that $|f(x)|<\varepsilon$ for all $x \in U_0$. Now $Y$ is compact and $U_0$ is open hence $Y\setminus U_0$ is closed hence compact. Now I am not able to proceed from here and also the $''\Rightarrow''$ direction.
Can you please give me some hint how to approach. Thank you.
($\Leftarrow$) To continue, observe that $\{\omega \in X: |f(\omega)|\geq \varepsilon\}\subset X\setminus U_0$.
($\Rightarrow$) Assume now that $f\in C(Y)$ and that $f$ (or more appropriately, its restriction to $X$) vanishes at infinity. We want to show that $f(0)=0$. Using continuity, it suffices to show that for each $\varepsilon>0$, there is some open neighborhood $U\subset Y$ of $0$ such that $|f(x)|<\varepsilon$ for all $x\in U$, $x\neq 0$.