I guess that your source was Wikipedia article on Sard's theorem which states at the end
In 1965 Sard further generalized his theorem to state that if $f:M\rightarrow N$ is $C^k$ for $k\geq \max\{n-m+1, 1\}$ and if $A_r\subseteq M$ is the set of points $x\in M$ such that $df_x$ has rank less than or equal to $r$, then $f(A_r)$ has Hausdorff dimension at most $r$.
Specializing to $m=n=1$ and $r=0$, we get the statement that the set of critical values of a $C^1$ function $f:\mathbb R\to\mathbb R$ has Hausdorff dimension $0$. (By the way, it looks strange when you write "at most $0$". The Hausdorff dimension is never negative, even for the empty set, according to its common definition.)
The above statement is false. Here is a counterexample.
Let $C$ be the ternary Cantor set. Define
$$f(x)=\int_0^x \operatorname{dist}(t,C)\,dt\tag1$$
where as usual, $\operatorname{dist}(t,C)=\inf\{|t-y|:y\in C\}$. You can visualize the graph of the function $t\mapsto \operatorname{dist}(t,C)$ as a collection of right isosceles triangles built on the gaps of $C$ (plus some uninteresting lines left and right of the set). Since $t\mapsto \operatorname{dist}(t,C)$ is continuous, $f\in C^1(\mathbb R)$.
Clearly, $f'(x)=0$ for every $x\in C$. I claim that
$$ |f(x)-f(y)| \ge \frac{1}{36}|x-y|^2,\quad x,y\in C \tag2 $$
Assuming (2) for now, we see that $f$ is injective on $C$, and that the inverse map $f^{-1}:f(C)\to C$ is Hölder continuous with exponent $1/2$. Recall that $\alpha$-Hölder maps do not increase the Hausdorff dimension by more than the factor of $1/\alpha$. Therefore, $\dim C\le 2\dim f(C)$; put another way,
$$\dim f(C)\ge \frac12 \frac{\log2}{\log 3} \tag3$$
It remains to prove (2). Take $x,y\in C$, $x<y$. Let $k$ be the smallest integer for which there is a gap of size $3^{-k}$ between $x$ and $y$. Then $x$ and $y$ are in the same component of the $(k-1)$ generation of pre-Cantor sets. Hence $y-x\le 3^{1-k}$.
Recalling the triangular shape of graph of $t\mapsto \operatorname{dist}(t,C)$ on each gap, we find that the area of the triangle is $\frac14 3^{-2k}$. Thus,
$$
f(y)-f(x)=\int_x^y \operatorname{dist}(t,C)\,dt \ge \frac14 3^{-2k} \ge \frac{1}{36}(y-x)^2
$$
as claimed in (2). $\quad \Box$
It is instructive to note that $f\notin C^2$ in this example. To make it $C^2$ smooth, I would have to replace triangles with something smoother, and also scale them differently in the vertical direction. The resulting $C^2$ map would still satisfy something like (2) but with a different power of $|x-y|$. Same goes for $C^k$ for all finite $k$; the larger $k$, the larger the exponent in (2), the smaller the lower bound in (3). But the bound will still be positive. Only for $C^\infty$ functions the counterexample disappears.
Now we are ready to appreciate the actual result that Sard proved in 1965, which was misquoted in Wikipedia. I quote from page 162 of Hausdorff Measure of Critical Images on Banach Manifolds, taking into account the errata. (Sadly, both links are paywalled.)
Define the function $c$ as follows:
$$c(k,\rho )= \begin{cases} 1 &\text{ if }\ k<\rho, \\ 2+k(k-\rho)/\rho &\text{ if }\ 0<\rho\le k, \end{cases} \qquad k=0,1, \dots, \ \rho>0$$
Suppose that $f$ is a $C^q$-map of a $C^q$-manifold $M$ of dimension $m <\infty$ with
countable basis into a $C^q$-Banach manifold $N$, where $1 \le q \le\infty$. Let $A_r$
denote the set of points of $M$ of rank $\le r$, $r= 0,1, \dots,m $. Put $k=m - r$.
Theorem 1. Suppose that $\rho > 0$. Then $f(A_r)$ is $(r + \rho)$-null if
$$q \ge c(k, \rho).$$
In order to conclude $\dim f(A_0)=0$ for a map $f:\mathbb R\to\mathbb R$, we must be able to use arbitrarily small $\rho>0$ above, and this requires $q=\infty$. On page 169 Sard spells this out as a corollary:
Corollary. If $f$ is a $C^\infty$-map, then $f(A_r)$ is $(r+ \rho)$-null for all $\rho > O$.
and adds:
This corollary, at least in the case in which $N$ is finite dimensional, is due to Dubovickii [8].
Indeed, А. Я. Дубовицкий published this result in 1962. His paper is in free access here.
