I have been hesitating on whether I should post this long solution, but I was unable to find better. I will just add missing details to the received answers and comments, which all conveyed the same idea on how to put a metric on the space of curves, but did not explain why it is a metric.

First, let me add one more statement from the book that I've quoted. Let $I$ be some fixed closed interval and let $X$ be a metric space. We will call a path $\gamma:I\to X$ *unstopping* if there is no open subintervals of $I$, where $\gamma$ is a constant (in the book they call such paths *never locally constants*, but in my opinion it is too mouthful). It is explained in the book that any path is a re-parametrization of an unstopping one, which together with the description quoted in the body of question gives that every curve is of the form $\left\{\gamma\circ\phi\left|\phi\in H\right.\right\}$, where $\gamma:I\to X$ is an unstopping representative and $H$ is the set of all increasing surjection from $I$ onto itself.

Let $G$ be the set of all strictly increasing bijections from $I$ into $I$. It is easy to see that $G$ is dense in $H$ with respect to the uniform topology, and since $I$ is compact, the correspondence $\phi\to\gamma\circ\phi$ is continuous from $H$ into $C\left(I,X\right)$ with respect to the uniform topologies. Thus $\left\{\gamma\circ\phi\left|\phi\in G\right.\right\}$ is dense in $\left\{\gamma\circ\phi\left|\phi\in H\right.\right\}$. Thus we can define the distance between the curves parametrized by unstopping paths $\gamma_{1},\gamma_{2}$ by $\widetilde{d}\left(\widetilde{\gamma_{1}},\widetilde{\gamma_{2}}\right)=\inf\left\{d\left(\gamma_{1},\gamma_{2}\circ\phi\right)\left|\phi\in G\right.\right\}$. Since $G$ is a group it is easy to show the triangle inequality. We are only left with proving non-degeneracy. We will need an additional concept for it.

For a unstopping path $\gamma:I\to X$ define $S_{\gamma}$ to be the collection of all closed subsets $J$ of $I$, which are components of $\gamma^{-1}\left(\gamma\left(J\right)\right)$. It is easy to show the following properties of this family.

**Lemma 1**

**(i)** $J\in S_{\gamma}$ if and only if it is a closed connected set, and each of the endpoints of $J$ belongs to $\overline{I\backslash \gamma^{-1}\left(\gamma\left(J\right)\right)}$, unless it is also an endpoint of $I$.

**(ii)** If $J_{1}, J_{2}\in S_{\gamma}$ and $J_{1}\bigcap J_{2}$, then $J_{1}\bigcap J_{2}\in S_{\gamma}$.

**(iii)** If $J_{1}, J_{2}\in S_{\gamma}$ and $\gamma\left(J_{1}\right)\subset \gamma\left(J_{2}\right)$, then either $J_{1}\subset J_{2}$, or $J_{1}\bigcap J_{2}=\varnothing$.

**(iv)** For any $t\in I$ the collection $\left\{int J\left|J\in S_{2},~t\in int J\right.\right\}$ forms a local base of the usual topology of $I$ at $t$. Also $\bigcap\limits_{t\in int J,~J\in S_{\gamma}}\gamma\left(J\right)=\left\{\gamma\left(t\right)\right\}$.

Fix two unstopping paths $\gamma_{1},\gamma_{2}$, such that $\widetilde{d}\left(\widetilde{\gamma_{1}},\widetilde{\gamma_{2}}\right)=0$. Since the uniform distance is at least the Hausdorff distance between the images, we get that that $\gamma_{1}\left(I\right)=\gamma_{2}\left(I\right)$. Without loss of generality, we can assume that $\gamma_{1}\left(I\right)=\gamma_{2}\left(I\right)=X$. Let us refine this last assertion. For $I_{1},I_{2}\subset I$ define $I_{1}\le I_{2}$ if $\sup I_{1}\le \inf I_{2}$, and the same for strict inequalities.

**Lemma 2**

For any sequence $\left\{I_{n}\right\}_{n=1}^{\infty}$ of closed subintervals of $I$ there is another sequence $\left\{J_{n}\right\}_{n=1}^{\infty}$, such that $\gamma_{1}\left(I_{n}\right)=\gamma_{2}\left(J_{n}\right)$ and whenever $I_{m}\subset I_{n}$ or $I_{m}\le I_{n}$, then $J_{m}\subset J_{n}$ or $J_{m}\le J_{n}$, respectively.

**Proof** There is a sequence $\left\{\phi_{k}\right\}_{k=1}^{\infty}$ such that $d\left(\gamma_{1},\gamma_{2}\circ\phi_{k}\right)\to 0$. Then the images of $\gamma_{2}\left(\phi_{k}\left(I_{n}\right)\right)$ converge to $\gamma_{1}\left(I_{n}\right)$, $\forall n$ in Hausdorff metric. Note that for each $k$ whenever $I_{m}\subset I_{n}$ or $I_{m}\le I_{n}$, then $\phi_{k}\left(I_{m}\right)\subset \phi_{k}\left(I_{n}\right)$ or $\phi_{k}\left(I_{m}\right)\le \phi_{k}\left(I_{n}\right)$, respectively because $\phi_{k}$ is increasing. Since the hyperspace of a compact $I$ is compact, $\left\{\phi_{k}\left(I_{n}\right)\right\}_{k=1}^{\infty}$ has a converging subsequence for each $n$. Using the Cantor's diagonal argument, we can find a subsequence $\left\{\phi_{l_{k}}\right\}_{k=1}^{\infty}$, and a sequence $\left\{J_{n}\right\}_{n=1}^{\infty}$ of closed subintervals of $I$ such that $\phi_{l_{k}}\left(I_{n}\right)\to J_{n}$, $\forall n$. Since "hypermap" is continuous, we have that $\gamma_{2}\left(J_{n}\right)=\lim\limits_{k\to\infty}\gamma_{2}\left(\phi_{l_{k}}\left(I_{n}\right)\right)=\gamma_{1}\left(I_{n}\right)$. **QED**

We can modify the previous lemma under the assumption that we are dealing with members of $S_{i}=S_{\gamma_{i}}$.

**Lemma 3**

For any sequence $\left\{I_{n}\right\}_{n=1}^{\infty}\subset S'_{1}$ there is another sequence $\left\{J_{n}\right\}_{n=1}^{\infty}\subset S'_{2}$, such that $\gamma_{1}\left(I_{n}\right)=\gamma_{2}\left(J_{n}\right)$ and whenever $I_{m}\subset I_{n}$ or $I_{m}< I_{n}$, then $J_{m}\subset J_{n}$ or $J_{m}< J_{n}$, respectively.

**Proof** We will add some auxiliary sets to our sequence, which will later serve as "boundaries". Assume that $I_{m}< I_{n}$ Then since $\max I_{m}\in\overline{I\backslash \gamma_{1}^{-1}\left(\gamma\left(I_{m}\right)\right)}$ and $\min I_{n}\in\overline{I\backslash \gamma_{1}^{-1}\left(\gamma\left(I_{n}\right)\right)}$, there
are $t,s\in\left[\max I_{m},\min I_{n}\right]$, such that $t\le s$, $\gamma_{1}\left(t\right)\not\in\gamma\left(I_{m}\right)$ and $\gamma_{1}\left(s\right)\not\in\gamma\left(I_{n}\right)$. Define $I_{n}^{m}=\left\{s\right\}$. Define $I_{m}^{n}=\left\{t\right\}$ and $I_{n}^{m}=\left\{s\right\}$. Note that $I_{m}\le I_{m}^{n}\le I_{n}^{m}\le I_{n}$.

The original sequence of sets together with all possible $I_{m}^{n}$ is still (at most) countable, so we can apply the previous lemma to it and get the collection $\left\{J'_{n}\right\}_{n=1}^{\infty}\bigcup \left\{J_{m}^{n}\left|I_{m}\bigcap I_{n}=\varnothing\right.\right\}$, which also has additional properties guaranteed by the lemma. Define $J_{n}$ to be the component of $\gamma_{2}^{-1}\left(I_{n}\right)$, which contains $J'_{n}$. Clearly, $J_{n}\in S_{2}$ and $\gamma_{1}\left(I_{n}\right)=\gamma_{2}\left(J_{n}\right)$.

If $I_{m}\subset I_{n}$, then $J'_{m}\subset J'_{n}$, and since $J_{m}, J_{n}\in S_{2}$, $\gamma\left(J_{m}\right)\subset \gamma\left(J_{n}\right)$ and $J_{m}\bigcap J_{m}\supset J'_{m}\ne\varnothing$, by part (iii) of Lemma 1 we conclude that $J_{m}\subset J_{n}$.

If $I_{m}< I_{n}$, then $J'_{m}\le J_{m}^{n}\le J_{n}^{m}\le J'_{n}$. Also $\gamma_{1}\left(I_{m}^{n}\right)=\gamma_{2}\left(J_{m}^{n}\right)$ is a single point, which does not belong to $\gamma_{1}\left(I_{m}\right)=\gamma_{2}\left(J_{m}\right)$. Hence $J_{m}<J_{m}^{n}$, and analogously $J_{n}^{m}<J_{n}$, from which $J_{m}<J_{n}$. **QED**

From the uniform continuity of $\gamma_{i}$ and previous lemma one can deduce that for any closed $Y\subset X$ which contains more than one point, the number of components $J$ of $\gamma_{1}^{-1}\left(Y\right)$, such that $\gamma_{1}\left(J\right)=Y$ is equal to the analogous number for $\gamma_{2}$.

Let $S'_{i}$ be the subcollection of $S_{i}$ consisting of sets, which are not singletons. Define a map $\phi:S'_{2}\to S'_{1}$ in the following way. Since $\gamma_{2}$ is unstopping, $Y=\gamma_{2}\left(J\right)$ is not a singleton for any $J\in S'_{2}$. Let $J_{1}< J_{2}< ...< J_{n}$ be all the elements of $S_{2}$ such that $\gamma_{2}\left(J_{k}\right)=Y$. Note that $J$ is among $J_{1}, J_{2}, ..., J_{n}$, say $J=J_{k}$. Let $I_{1}< I_{2}< ...< I_{n}$ be all the elements of $S_{1}$ such that $\gamma_{1}\left(I_{k}\right)=Y$. Define $\phi\left(J\right)=I_{k}$.

If $J^{1}, J^{2}\in S'_{2}$ and $J^{1}\subset J^{2}$, then $\phi\left(J^{1}\right)\subset \phi\left(J^{2}\right)$. Indeed if $J^{1}_{1}< J_{2}^{1}< ...< J_{n}^{1}$ are all the elements of $S'_{2}$ such that $\gamma_{2}\left(J^{1}_{k}\right)=\gamma_{2}\left(J^{1}\right)$ and $J^{2}_{1}< J_{2}^{2}< ...< J_{m}^{2}$ are all the elements of $S_{2}$ such that $\gamma_{2}\left(J^{2}_{k}\right)=\gamma_{2}\left(J^{2}\right)$, the statement follows from applying Lemma 3 to the collection of $J_{k}^{j}$.

This last property allows us to extend $\phi$ to the singletons, or with some abuse of notations, to $I$. Namely, for $t\in I$ define $\phi\left(t\right)=\bigcup\limits_{t\in int J,~J\in S_{2}}\phi\left(J\right)$. From the properties of $S_{2}$ and $\phi$, we get that the collection $\left\{\phi\left(J\right)\left|t\in int J,~J\in S_{2}\right.\right\}$ has finite intersection property, and so it has a nonempty intersection. Moreover, since $\gamma_{2}\left(J\right)=\gamma_{1}\left(\phi\left(J\right)\right)$ for any $J\in S'_{2}$, we get that $\gamma_{1}\left(\phi\left(t\right)\right)\subset\bigcup\limits_{t\in int J,~J\in S_{\gamma}}\gamma_{2}\left(J\right)=\left\{\gamma_{2}\left(t\right)\right\}$, and so $\phi\left(t\right)$ is a singleton. Moreover, $\gamma_{1}=\gamma_{2}\circ\phi$. Thus our goal is accomplished once we prove the following lemma.

**Lemma 4**

$\phi$ is a strictly increasing surjection into $I$.

**Proof** From the symmetry it is easy to see that $\left\{\phi\left(J\right)\left|t\in int J,~J\in S_{2}\right.\right\}=\left\{J\left|\phi\left(t\right)\in int J,~J\in S_{1}\right.\right\}$, and so $\phi$ is invertible as a map from $I$ into itself; it is also easy to see that the endpoints of $I$ are fixed points of $\phi$.

Finally since the families $\left\{int J, J\in S_{i}\right\}$ form a base of topology of $I$, for each $i=1,2$, it follows that $\phi$ is continuous. Indeed, for any open neighborhood $U$ of $\phi\left(t\right)$ for $t\in I$ there is $J\in S_{1}$ such that $t\in int J$ and $J\subset U$, and then $\phi^{-1}\left(J\right)$ is an open neighborhood of $t$, such that $\phi\left(s\right)\in\phi\left(\phi^{-1}\left(J\right)\right)=J\subset U$, for any $s\in\phi^{-1}\left(J\right)$. **QED**