If $x \ne y$ in a metric space $M$, how can I show that there are disjoint open sets $U,V$ with $x \in U$ and $y \in V$?
The question goes on with $U$ and $V$ can be chosen so that $\bar U$ and $\bar V$ are disjoint
I honestly have no idea how to approach this.
On
If $x\neq y$, then $s=d(x,y)>0$. If $t<\frac s 2$, then you can show that the open $t$-balls around $x$ and $y$ are disjoint.
The closures of open $t$-balls $\{ z\in M : d(x,z)<t\}$ are contained in the closed $t$-balls $\{z\in M: d(x,z)\le t\}$. So if you choose $t<\frac s2$, the closed $t$-balls around $x$ and $y$ will be disjoint (proof: if $z$ is in both, then $d(x,z)\le t$ and $d(y,z)\le t$ so $s=d(x,y)\le t+t<s$, contradiction) and so the closures of the open $t$-balls will be disjoint as well.
Hint: Draw two dots on a piece of paper. Draw a small circle of the same size around each dot, centered at the dot, being sure the circles are small enough that they don't meet.
In terms of the distance between the two dots, how small (of what maximum radius) do the circles have to be so that they don't meet?