How to prove that no two-to-one function can be continuous? This is question

I have almost done this question, but why can we only assume $a<c<d\leq b$ and why not $a\leq c<d<b$ and also the case when $a=c$ and $b=d$. That is my professor asked me, could you explain for me?  

 A: The point is that since at least two of $c, d, e, f$ are in the interior of $[a, b]$, we must have:


*

*EITHER at least one of $c, d$ is in the interior of $[a, b]$, 

*OR at least one of $e, f$ is in the interior of $[a, b]$.
(Of course, perhaps both possibilities hold.)
Each possibility further refines into two more possibilities: e.g., if we have that at least one of $c, d$ is in the interior of $[a, b]$, then


*

*EITHER $c$ is in the interior of $[a, b]$ - in which case $a<c<d\le b$ -

*OR $d$ is in the interior of $[a, b]$ - in which case $a\le c<d<b$.
(Again, perhaps both possibilities hold.)
So we know that (at least) one of the following four cases holds:


*

*$a<c<d\le b$,

*$a\le c<d<b$,

*$a<e<f\le b$, or

*$a\le e<f<b$.
The point is that each of these cases is handled in the same way, so we only have to explicitly write out one of them.
That said, you should understand why they are each handled in the same way - look at the proof of the "$a<c<d\le b$" case, and see why $(i)$ nothing special about "maxima" (vs. "minima") is used, and $(ii)$ why nothing special about the left extremum being in the interior (vs. the right extremum being in the interior) is used.
