Proving that $\overline{([a,b],\le)}=\overline{([c,d],\le)}$, i.e., the order types of two closed intervals are the same 
Prove that for all $a,b,c,d : a\le b  : c\le d$ we have
1.$\overline{[a,b]}=\overline{[c,d]},$
2.$\overline{(a,b)}=\overline{(c,d)}$
3.$\overline{[a,b)}=\overline{[c,d)}$
4.$\overline{(a,b]}=\overline{(c,d]}$
About the notation: $\overline{[c,d]}=\overline{([c,d],\le)}$, it's an order type, like $\omega=\overline{(\mathbb N,\le)}$, I also saw in some book they wrote it like so: $\langle[c,d],\le\rangle$.

I don't know if one interval covers the other so I think the right approach would be to show WLOG that every interval say $[a,b]$ is isomorphic to $[0,1]$, so $a\to 0, \ b \to 1$ and the function would be a straight line where $\frac {a-b}{1-0}$ is the slope and $a$ intersects the $y$ axis, so the function is: $f(x)=(a-b)x+a$.
This function should work for all of the above intervals I think. Is my approach correct ?

We can assume that $a < b$ and $c < d$, as otherwise we get degenerated spaces (singletons or empty sets, for which it's evident that they have the same order type...)
Use $f(x) = (b-a)x + a$ from $[0,1]$ to $[a,b]$ instead. It's increasing as $b-a > 0$, and it will also work for the other types of interval. And we use that having the same order type is an equivalence relation, of course. 
