Theorem: (Sard) Let $f:U\to \mathbb R^p$ be a smooth map with $U$ open in $\mathbb R^n$, and let $C$ be the set of all critical points of $f$. Then $m([f(C)]=0$ where $m$ is the Lebesgue measure of $\mathbb R^p$.
A proof goes by induction on $n\leq 0$, while $p>0$. The base case is obvious. Let $C_k$ denote the elements of $U$ for which the partial derivatives of $f$ up to the $k$-th order vanish at the said element. We may brake the proof in three steps. We will be focusing on the first step.
1) $m([f(C-C_1)])=0$
2) $m([f(C_k-C_{k+1})])=0, \quad k>0$
3) $m([f(C_k)])=0, \quad\forall i>n_0$, for some natural number $n_0$.
Step 1) Assume $p>1$, since $C=C_1$ when $p=1$. For each $y\in C-C_1$ we will find a an open neighborhood $V$ containing $y$ for which $m([f(V\cap C)])=0$.
Up to an isometry, $y\not\in C_1\Rightarrow \frac{\partial f_1}{\partial x_1}(y)\not=0$. Consider $h(x):=(f_1(x),x_2,x_3,\dots,x_n)$. Since $dh_y$ is nonsingular, $h$ maps some neighborhood of $y$ diffeomorphicaly onto an open set $V'$. The composition $g=f\circ h^{-1}$ maps $V'$ to $\mathbb R^p$, etc.
Questions: How can we see that $g(t,x_2,\dots,x_n)\in t\times\mathbb R^{p-1}$, for each $(t,x_2,\dots,x_n)\in V'$? How can we conclude from the preceding that $g$ caries hyperplanes to hyperplanes?
Comment: Obviously, the first component of $g(t,x_2,\dots,x_n)$ is $t$, but there are many ways to include $\mathbb R^{p-1}$ in $\mathbb R^p$. I guess that the meaning of the statement is that some suitable inclusion exists but it varies with $t$.
Continuing: A point is critical for $g^t:=g|_{t\times\mathbb R^{n-1}\cap V'}$ if and only if it is critical for $g$. Therefore, the set $C'$ of critical values of $g$ intersects intersects each hyperplane $t\times\mathbb R^{p-1}$ in a set of measure zero. By a theorem of Fubini, $m([g(C')])=0$ but $f(V\cap C)=g(C')$. Hence, $m([f(C-C_1)])=0$ because it is a countable union of sets of measure zero. q.e.d.
Questions: Again many ways exist to restict $g$ to a hyperplane with a fixed first coordinate. Which restriction may we mean? (I think that we must mean every possible restriction.)
Moreover, the theorem of Fubini states that if a set $A\subset\mathbb R^n=\mathbb R\times\mathbb R^{n-1}$ intersects each hyperplane $\text{const}\times\mathbb R^{n-1}$ in a set of measure zero, then $A$ must also have measure zero. It seems that we have not satisfied the conditions of the theorem. In order to satisfy them we need to argue about every possible hyperplane.
In John Milnor's book, "Topology from the Differentiable Viewpoint", there is a proof of Sard's theorem.
We have a map $g:f\circ h^{-1}$
Where $f:U \to R^p$ where U is an open set in $R^n$
$h:U\to R^n$ where $h(x)=(f_1(x),x_2,...,x_n)$
$h$ is a diffeomorphism which means its a bijection
For each point $(t,x_2,...x_n)\in V'$,
Let $h^{-1}:V' \to R^n$ where $h^{-1} = (t,x_2,...,x_n)$
Then $f(t,x_2,...,x_n)=(t,x_2,...,x_{p-1})$ which is a hyperplane
Therefore the composition $g:f\circ h^{-1}$ maps hyperplanes to hyperplanes