In an interval $[x,y]$ in a poset, with $x\neq y$, there is only one chain of length one. It is the chain $x<y$. Since the chain must begin with the element $x$ and end with the element $y$, there are no other options.
As an example let’s compute the Mobius value $\mu(a,d)$ in the chain $a<b<c<d$.
First, there is no chain of length zero beginning at $a$ and ending at $d$, so $C_0(a,d)=0$. There is one chain of length one, namely $a<d$, so $C_1(a,d)=1$. There are two chains of length two, $a<b<d$ and $a<c<d$, so $C_2(a,d)=2$. Finally, there is one chain of length three, i.e. $a<b<c<d$, so $C_3(a,d) = 1$.
Using Hall’s theorem we can now compute:
$$\mu(a,d) = C_0 -C_1 + C_2 -C_3= 0-1+2-1 =0$$