The answer is *no* to the first question and *yes* to the second one. 

Basically, not every part of the Osgood curve is an Osgood curve by itself, because of the "joins". The standard Osgood curve is constructed as the limit of a nested sequence of sets $C_n$, where $C_n$ is a finite union of squares connected with straight segments ("joins"). Once such a "join" appears in $C_k$ for some $k\in\mathbb N$, it will remain 
in every $C_n$ with $n\geq k$, and thus will be a part of the Osgood curve. The 2D Lebesgue
measure of a "join" is zero, of course. 

The first example of the Osgood curve with the property that its every sub-arc is also an Osgood curve was provided by Sierpinski. His construction was later improved by Knopp. The versions of the Osgood curve by Sierpinski and Knopp are obtained as the limits of sequences of polygons (without any "joins"). Knopp actually constructed a one-parametric family  of Jordan curves of positive 2D Lebesgue measure $\lambda$ for any $\lambda\in(0,1)$. The limiting cases correspond to the fractal von Koch curve ($\lambda=0$) and the space-filling Sierpinski-Knopp curve ($\lambda=1$).   


*Space-filling curves* by Hans Sagan contains a good survey of the results.

**Edit added:** You may also watch a few steps of the construction of Knopp's Osgood curve 
[here][1].

![alt text][2]


  [1]: http://demonstrations.wolfram.com/KnoppsOsgoodCurveConstruction/
  [2]: http://demonstrations.wolfram.com/KnoppsOsgoodCurveConstruction/HTMLImages/index.en/popup_5.jpg