Rudin states that definition of 1-surface as the same as definition 6.26.
For example, in definition 10.10 we can take $D=[0,1]\cup [2,3]$ which is compact in $\mathbb{R}^1$. If $F$ is $C'$ - mapping then $F:D\to E$ is 1-surface, but it does not coincides with definition of curve (6.26).
Can anyone explain this moment please? Maybe I misunderstand smth.
There is a remark between the two parts you quoted:
We shall confine ourselves to the simple situation in which $D$ is either a $k$-cell or the $k$-simplex $Q^k$ described in Example $10.4$. The reason for this is that we shall have to integrate over $D$, and we have not yet discussed integration over more complicated subsets of $R^k$. It will be seen that this restriction on $D$ (which will be tacitly made from now on) entails no significant loss of generality in the resulting theory of differential forms.
It is this restriction that makes $1$-surfaces and continuously differentiable curves the same. For $k = 1$, $1$-cells are closed intervals, and the $1$-simplex $Q^1 = [0,1]$ is a particular closed interval.
Knowing that a continuously differentiable map defined on a closed interval $D$ can be extended to a continuously differentiable map defined on an open neighbourhood $U$ of $D$, we see that indeed each continuously differentiable curve is a $1$-surface, and the restriction on $D$ conversely makes every $1$-surface a continuously differentiable curve.
Without that restriction, your example gives a $1$-surface that is a union of finitely many continuously differentiable curves. But without that restriction, even more complicated situations would be allowed, we could choose for example a Cantor set for $D$ and get a $1$-surface that doesn't look curve-like.
