Does a path always look like a path? My question is simple. I always wonder if it could be possible the trace of a path $\gamma:\mathbb R\to \mathbb R^3$ be a solid figure. If don't, why not? 
 A: It is possible, yes. There are things called "space-filling curves" which would do the trick.
The prototypical example is a particular map $\gamma : [0,1] \to [0,1]^2$ which is both continuous and surjective (though not differentiable or invertible, obviously). You can extend this to get a continuous surjective map $\gamma_n : [0,1] \to [0,1]^n$ pretty easily then.
Using this then, you could map the unit cube to any volume or surface of your choosing in $\mathbb{R}^3$ to get what you're looking for.
A: That's depends on the definition of "path". There are definitely surjective functions from $\mathbb R$ to $\mathbb R^3$, but the question is if they would fit your definition of "path".
You can for example construct a continuous function from $\mathbb R$ to $\mathbb R^3$, by first enumerating unit cubes of $\mathbb R^3$ and then construct a function that jumps from cube to cube at regular interval. This shows that you only have to be able to construct a cube filling function from a finite interval to a unit cube. 
Basically what one do is divide the cube in eight and have a curve passing through all parts, then you divide the cube again and have a curve passing through all 64 parts. Doing this carefully you'll end up with a series of continuous functions that converges uniformly and therefore it's limit is continuous, but also visits all points in the cube. 
To get an idea you could check out the Peano curve
If you on the other hand require the curve to be bijective I'd guess you're out of luck. You can't do that in two dimensions at least due to Jordan curve theorem (i think).
