A set that is open in any metric space that contains it 
Let $X$ be a set with the following property: For all metric space $Y$ such that $X\subset Y$ we have that $X$ is an open set on $Y$. $X$ should be the empty set?

 A: The question makes sense only if $X$ is metrizable, in which case the answer is yes. 
If $X\ne\varnothing$, fix $p\in X$. Let $Y_0=\{y_n:n\in\Bbb N\}$ be a set of distinct points not in $X$, and let $Y=X\cup Y_0$. Let $d$ be a metric on $X$, and define a metric $d_1$ on $Y$ as follows. 
$$d_1(x,y)=\begin{cases}
d(x,y),&\text{if }x,y\in X\\
2^{-n},&\text{if }\{x,y\}=\{p,y_n\}\\
|2^{-n}-2^{-m}|,&\text{if }\{x,y\}=\{y_n,y_m\}\\
2^{-n}+d(x,p),&\text{if }x\in X\setminus\{p\}\text{ and }y=y_n\\
2^{-n}+d(y,p),&\text{if }y\in X\setminus\{p\}\text{ and }x=y_n\;.
\end{cases}$$
It’s not hard to check that $d_1$ is a metric on $Y$ that agrees with $d$ on $X$. However, $p\in\operatorname{cl}_YY_0$, so $X$ is not open in $Y$. (In case the details obscure what’s really going on, I’ve just added a simple sequence $\langle y_n:n\in\Bbb N\rangle$ converging to $p$.)
Added: The same idea works in general topological spaces. Just declare a set $V\subseteq Y$ to be open iff either it’s an open subset of $X$ that does not contain $p$; it’s a subset of $Y_0$; or $p\in V$, $V\cap X$ is open in $X$, and there is an $n_0\in\Bbb N$ such that $V\supseteq\{y_n\in n\ge n_0\}$. (If you’re familiar with quotient spaces and the one-point compactification, this $Y$ is homeomorphic to the quotient of $X\sqcup Y_0^*$, where $Y_0^*$ is the one-point compactification of the discrete space $Y_0$, obtained by identifying $p$ and the point at infinite in $Y_0^*$.)
