Topological characterization of the closed interval $[0, 1]$. I would like to learn purely topological characterizations of the closed real intervals (to justify the existence of algebraic topology). In particular, such a characterization should not use real numbers. I would like to see in what sense $[0, 1]$ is topologically more important than, for example, the set of rationals, or the complex unit disc. I am especially interested in why the unit interval is so important in the study of compact Hausdorff spaces (metrization, Urysohn's lemma).
I have come up with the following one.
Start with a definition of path connectedness that does not use $\mathbb R$: $x_1$ and $x_2$ are connected by a path in $X$ if for every Hausdorff compact $C$ and $a,b\in C$, there is a continuous $f\colon C\to X$ such that $f(a)=x_1$ and $f(b)=x_2$.  Now, if $X$ is a Hausdorff space and distinct points $x_1$ and $x_2$ are connected by a path in $X$, then there is a minimal subspace of $X$ in which $x_1$ and $x_2$ are still connected by a path, and every such subspace is homeomorphic to $[0, 1]$.
Roughly speaking, $([0, 1], 0, 1)$ is the minimal bi-pointed Hausdorff space such that every bi-pointed Hausdorff compact can be mapped into it with the distinguished points being sent onto the distinguished points.
Maybe in some sense it can be also said that $X = [0, 1]$ is the "minimal" Hausdorff space such that every Hausdorff compact embeds into $X^N$ for some $N$, but i do not know how to make this precise.
My question is: what are other "natural" topological characterizations of $[0, 1]$?

Update:  I have duplicated this question on MathOverflow.
 A: 
This is a duplicate of my answer on MO

Consider the class of all Hausdorff compacts with distinct points (i.e. which have more than $1$ point) that are absolute retracts in the class
of Hausdorff compacts.
Then $[0,1]$ is up to homeomorphism the only member of this class that embeds into every other.
A: A lesser-known characterization than the cut-point/end-point one is: A compact, connected metric space ('continuum') is an arc if and only if it is arc-like and path-connected.
A continuum $X$ is arc-like if for every $\epsilon > 0$ there is a continuous mapping $f: X \rightarrow [0,1]$ with diam$(f^{-1}(y)) < \epsilon$ for all $y \in [0,1]$.
There are also characterizations within the setting of locally connected spaces.  The book 'Continuum Theory' by Sam Nadler has a lot of characterizations of the arc.
A: It is the unique second countable continuum with precisely two non-cutpoints. This is due to Veblen, according to this overview.
A continuum is a connected and compact Hausdorff space, and a cutpoint (in a connected space) is a point that when removed leaves the remainder disconnected. The second countable is a (non-reals using) way of saying metrisable. One can prove the compactness and connectedness purely from order completeness of the order.
A: In my opinion all characterizations of the interval $[0,1]$ that are not using the totality of real numbers are far from being natural and are awkward at least.
It is very simple to define what a closed interval is when you have the notion of the real numbers: it is a compact and connected subspace. The other way, from the interval to the whole totality of real numbers, is unnatural.
You can of course argue that you do not need the totality of real numbers to define the closed interval. It is true. But you still need the concept of the totality of real numbers (at least on the intuitive, naive level) to convince yourself that the space you defined really is (isomorphic) to the closed interval. Without it (assume you never heard about the set of real numbers) you just have defined some species of topological space. You do not know apriori that it has all the properties you demand from the closed interval, since you don't know what a closed interval is. Therefore you do not know that it is useful from your perspective.
From the philosophical point of view such a definition is a classical vicious circle. 
Do you really want to justify the existence of algebraic topology on the basis of vicious circle?
Please stay with what is simple, beautiful and not awkward.
Alexey, in your comment you asked about properties making $[0,1]$ more significant than other spaces. I believe that the properties making $[0,1]$ (and its higher dimensional analogues) so special are not of purely topological nature. 
Algebraic topology is about functors from the topological world to the algebraic world. Those functors are used to extract topological knowledge from algebraic relations. From the topological perspective the functors of algebraic topology have to be computable to be of any value. For instance: you want to prove that two spaces are not homeomorphic. You are computing their fundamental groups (or other algebraic functors) and see that they are not isomorphic. The reason that you succeded was beacause you were able to compute those algebraic objects effectively. In this toy example I would like to point out that the crucial property of the functors of algebraic topology is computability.
To make my point of view more precise: usefulness and significance of the intervals in the realm of algebraic topology are not of purely topological nature. 
They are of topological, algebraic and computational nature. 
Just like the algebraic topology is about topology, algebra and computation of functors.
The intervals are so special because they can be used to construct functors that are "easily" computable. (I don't think that this reason can be dissected any further. Also, it does not exclude the possibility that other spaces have this property.)
