# Continuous path contains curve without loops

I'm trying to prove the Lemma 3.1 from Falconer's geometry of fractal sets:

Let $\psi : [a,b] \to \mathbb{R}^n$ be a continuous mapping with $\psi(a) \neq \psi(b)$. Then $\psi[a,b]$ contains a curve joining $\psi(a)$ to $\psi(b)$.

In this statement, by a curve it is meant a continuous injective mapping. The proof provided in the book seems to be wrong and I had no success in fixing the argument. The proof goes along the following lines:

For every multiple point $x \in \psi[a,b]$ define the interval $I_x = [t_1, t_2]$ as the maximum interval such that $\psi(t_1) = \psi(t_2) = x$. Notice that whenever we have $I_y \subset I_x$ we can 'discard' the interval $(t_1, t_2)$ from the original curve getting a new continuous mapping in which both $x$ and $y$ are no longer multiple points.

My main difficulty is manage to apply this shortening globally. In the book's proof it is defined the family $\mathcal{I}$ of intervals $I_x$ that are contained in no others and it is claimed that this family is composed of countably many disjoint proper closed intervals. This claim is false. Any ideas?