In the book Topology from the differential viewpoint - Milnor, there is a proof of Sard Theorem : When $f:X^{n+m}\rightarrow Y^n$ is a smooth map where $X^N$ has $N$ dimension, then $f(C^1)$ has a measure $0$ where $C^1 =\{ x\in X |df_x=0\}$.
Here $C^1$ is a set of isolated points ? Note that $C^1$ is an intersection of $(n+m)n$ equations $\frac{\partial f_i}{\partial x_j}=0$. Here $C^1$ can be a curve or a surface ?
Not necessarily is $C_1$ a set of isolated points. Just consider the example of a constant function $f: \mathbb{R}^2 \to \mathbb{R}$. It is $df_{(x,y)} \equiv 0$, so $C^1 = \mathbb{R}^2$, a surface. If you want $C^1$ to be a curve, consider again any constant function $f: \mathbb{R} \to \mathbb{R}$, whose $C^1$ is $\mathbb{R}$, a curve.
Another example is the function $g:\mathbb{R}^2 \to \mathbb{R}$, $g(x,y) = (x-y)^2$, which satisfies $df_{(x,y)} = 0$ if and only if $x =y$. This means that for the function $g$, $C^1 = { (x,y)\in \mathbb{R}^2 : x= y}$, which is a curve in $\mathbb{R}^2$.