how to choose which one holds? let $X$ be any set with the property that for any two metrics $d_1$ and $d_2$ on $X$, the identity map 
$id:($X$,d_1$)$\to ($X$,d_2)$ is Continuous.
which of the following are true?
1) $X$ must be a singleton.
2) $X$ can be any finite set.
3) $X$ cannot be finite.
4) $X$ may be infinite but not uncountable.
how to solve this? help me
 A: Hints:
1) Every metric space is Hausdorff
2)Every finite Hausdorff space has the discrete topology.
Can you choose the right option now? Let me know if you need me to elaborate. 
Added: To rule out $4$, consider the metric $d_1$ inherited by $\mathbb Z$ by embedding it in $\mathbb R$ by the function $\phi$, where $\phi(n)=\frac{1}{n}$ for $n\neq 0$ and $\phi(0)=0$.   Let $d_2$ be the usual metric of $\mathbb Z$.   The singleton $\{0\}$ is an open set in $(\mathbb Z,d_2)$ but it is not open in $(\mathbb Z,d_1)$ because every interval around $0$ contains infinitely many terms of the form $\frac{1}{n}$. So the identity map from $(\mathbb Z,d_1)$ to $(\mathbb Z,d_2)$ is not continuous.
A: It's easy if you know that all the norms over $\mathbb{R}$ are equivalent. Then you function is continuous if you choose $X=\mathbb{R}$ and $d_1$ $d_2$ metrics induced by norms over $\mathbb{R}$. In fact $ \exists c \in \mathbb{R}: d_2(x,y) \le c \ d_1(x,y)$ and if you take a succession $(x_n)$ of real numbers convergent to $x$ for the metric $d_1$ you have 
$$ d_2(id(x_n),id(x))=d_2(x_n,x) \le c \ d_1(x_n,x) \rightarrow 0 $$ then $id$ is continuous. This answers 1,3,4.
For 2: it's true because if a succession $(x_n)$ tends to $x$ in a finite set then it's definitively equal to $x$, meaning that definitively $d_1(x_n,x)=0$. But Then also $d_2(id(x_n),(x))=d_2(x_n,x)=0$ definitively.
