Prove that S is the disjoint union of intervals, whose total length is at least 1. 
Let $I$ be a set of an odd number of closed unit intervals on the real
  line, and let $S$ be the set of points which are covered by exactly an
  odd number of elements of $I$.
Prove that $S$ is the disjoint union of
  intervals, whose total length is at least 1.

My solution is to prove it by induction on the number of intervals the points come from.
$\textit{BASIS STEP}:$
Let us assume the points in $S$ are selected from just one interval of $I$. This means $S = I_n$, so clearly $S$ is a disjoint union. Also, each member of $I$ is guaranteed to be of unit length so $S$ also has at least length 1. This completes the basis step.
$\textit{INDUCTIVE STEP}:$
Suppose the claim holds for the sets whose points come from $2n+1$ intervals of $I$, $n\geq 1$. We want to show this is also the case for the sets whose points belong to $2n+3$ intervals of $I$.
For this purpose, select points from exactly $2n+3$ intervals of $I$ to construct $S$. Now remove the interval that has the minimum starting point, and also the interval with the maximum ending point. The resulting set is composed of $2n+1$ intervals of $I$.
By the induction hypothesis, ... .Then I'm gonna use the subtracted intervals to show that we can still form a disjoint union, with a length of at least 1.
Okay I am still working on this proof. I want to ask if I can really prove it by induction? Am I on the right path? 
 A: Let $S$ be a set of points covered by exactly $2(k+1)+1=2k+3$ elements in $I$, i.e. $S=\bigcup_{i=1,\dotsc,2k+3}I_i$. Call the interval containing the minimum point $I_{2k+2}$, and the one containing the maximum $I_{2k+3}$. Let $T$ be a set of points covered by exactly $2k+1$ elements, described in the BIG brackets here:
$$S=\overbrace{[S\setminus\{T\cup I_{2k+3}\}\bigg[\dotsc]}^{I_{2k+2}} 2k+1\ \text{closed unit intervals}\overbrace{[\dotsc\bigg]S\setminus\{T\cup I_{2k+2}\}]}^{I_{2k+3}}.$$
Note, we are not excluding the intervals that have the minimum and maximum points, but the disjoint parts of each (since they may overlap as described in the picture).
Call $S\setminus\{T\cup I_{2k+3}\}$ the LDI (left-disjoint interval); and likewise RDI for $S\setminus\{T\cup I_{2k+2}\}$. By induction hypothesis, $T$ is a disjoint union of intervals, of total length $\ge1$. Then of course $T\cup\mathrm{LDI}\cup\mathrm{RDI}=S$ is a disjoint union of intervals, of total length $\ge1$.
Please tell me if there are any mistakes.
