There are really two separate questions here: can you define the unit interval space without talking about real numbers, and can you define path-connectedness without talking about the unit interval space? The answer to both is yes; let me address the second question first.
Let $P$ be a topological space and let $a,b\in P$ be two points. Say that a space $X$ is $(P,a,b)$-connected if for any $x,y\in X$, there is a continuous map $f:P\to X$ such that $f(a)=x$ and $f(b)=y$. Of course, for $(P,a,b)=([0,1],0,1)$, this is just the usual definition of path-connectedness.
However, there is a more "universal" characterization of path-connectedness that doesn't require you to know about the space $[0,1]$. Namely, a space $X$ is path-connected iff it is $(P,a,b)$-connected for all compact Hausdorff spaces $P$ with two distinct points $a,b\in P$.
To prove this, suppose $X$ is path-connected, $P$ is a compact Hausdorff space, $x,y\in X$, and $a,b\in P$ are distinct. Since $X$ is path-connected, there is a path $g:[0,1]\to X$ such that $g(0)=x$ and $g(1)=y$. By Urysohn's lemma, there is a continuous map $h:P\to [0,1]$ such that $h(a)=0$ and $h(b)=1$. The composition $gh:P\to X$ is then continuous and satisfies $g(a)=x$ and $g(b)=y$.
The idea here is that you could use any space $P$ with two chosen points $a$ and $b$ to define a notion of "paths" in a space. However, if you restrict to compact Hausdorff spaces $P$, then the ordinary interval $[0,1]$ is the "strongest possible kind of path" you can have in a space: if you have a $[0,1]$-path between two points, then you have a $P$-path for every other compact Hausdorff space $P$ as well.
(Of course, all we used about compact Hausdorff is that we know there is a map $P\to [0,1]$ separating $a$ and $b$. However, I phrased everything in terms of the compact Hausdorff condition since this is a natural condition you can define without already knowing about the space $[0,1]$.)
OK, now let me say a little about the first question. There are in fact many different ways to uniquely characterize the space $[0,1]$ up to homeomorphism without reference to the reals or anything that is essentially equivalent to constructing the reals. In fact, you can deduce one from the answer I gave to the second question above.
Namely, say that a compact Hausdorff space $(P,a,b)$ equipped with two distinct points is a universal path if it has the special property of $[0,1]$ noted above: whenever there is a $(P,a,b)$-path between two points $x$ and $y$ of an arbitrary space, there is also a $(Q,c,d)$-path from $x$ to $y$ for any compact Hausdorff space $Q$ with two distinct points. Say that a universal path $(P,a,b)$ is minimal if for any other universal path $(Q,c,d)$, there is an embedding $P\to Q$ sending $a$ to $c$ and $b$ to $d$.
I now claim that $([0,1],0,1)$ is the unique minimal universal path (up to homeomorphism). We know it is universal. To show that it is minimal, let $(Q,c,d)$ be any universal path. Since the identity map $Q\to Q$ is a $(Q,c,d)$-path from $c$ to $d$ in $Q$, universality implies there is a $([0,1],0,1)$-path from $c$ to $d$ in $Q$. But if there is a path between two points of a Hausdorff space, there is also a path which is an embedding (see Does path-connected imply simple path-connected?). Thus there is an embedding $[0,1]\to Q$ sending $0$ to $c$ and $1$ to $d$.
Now suppose $(P,a,b)$ is any minimal universal path. The previous paragraph shows that there is a $([0,1],0,1)$-path from $a$ to $b$ in $P$. Now since $([0,1],0,1)$ is a universal path, minimality of $P$ says that $P$ embeds in $[0,1]$ sending $a$ to $0$ and $b$ to $1$. But since $P$ contains a path from $a$ to $b$, the image of this embedding contains a path from $0$ to $1$, and thus contains all of $[0,1]$. Thus the embedding is actually a homeomorphism $P\to [0,1]$.
As I mentioned, this is just one of many ways of characterizing $[0,1]$. For another characterization that also relates closely to the intuitive notion of "paths", see this answer by Tom Leinster on MO.
