Let $(X,d)$ be a metric space and $G \subseteq X$ be a $d$-open set. For any $g \in G$, is the set $G\setminus\{g\}$ $d$-open in $X$? Hello stackexchange!
I would like to tell you my ideas about this question:
Firstly I tried to find a counter-example.
If $X = N$ and $d(m, n)$ = |m-n|,
then $$B_{\frac{1}{2}}(1) = \{m \in N : |m-1| < \frac{1}{2} \}$$
Since $\{1\}$ is the only element of the set $\{1\}$ and $B_{\frac{1}{2}}(1) = \{1\} \subseteq \{1\}$, the set $\{1\}$ is open. So if we remove this single point from the set we will obtain $\emptyset$. Also, we know that an empty set in any metric space is clopen. It means it's also closed. Can it be a correct answer for this question?
I'm open for any small pieces of hints!
Thanks!
 A: In a metric space, $\{g\}$ is always closed because it is the intersection of all $\overline{B(g,\varepsilon)}$ for $\varepsilon >0$. Then, $X\setminus \{g\}$ is open. As the intersection of two open subsets of $X$ is open, so is $G \setminus \{g\} = G \cap \left(X \setminus \{g\}\right)$.
A: 
Firstly I tried to find a counter-example.
(...)
Also, we know that an empty set in any metric space is clopen. It means it's also closed. Can it be a correct answer for this question?

I'm not sure what your thought is here. Sure, $\emptyset$ is closed. But that's not relevant. What is relevant is that it is open. And hence this is not a counterexample.
In metric spaces a subset $U$ is open if and only if for any $x\in U$ there is $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$.
Assume that $G$ is open and consider $G\backslash\{g\}$. Take any $x\in G\backslash\{g\}$. We want to show that some open ball $B(x,\epsilon)$ is fully contained in $G\backslash\{g\}$. Since $x\in G$, which is open, then there is $\tau >0$ such that $B(x,\tau)\subseteq G$. Now let $\epsilon=\min\{\tau,d(x,g)\}$ and note that $\epsilon>0$ and $B(x,\epsilon)\subseteq B(x,\tau)\subseteq G$. But since $\epsilon\leq d(x,g)$ then $g\not\in B(x,\epsilon)$. Meaning $B(x,\epsilon)\subseteq G\backslash\{g\}$. Which completes the proof.
