I think the pages 249-250 are the most relevant source in ariane's pdf. Sierpinski outlines how to go from the cardinality hypothesis to the existence of a non-measurable set, as per Ashutosh's précis, although without specific reference to the Lebesgue density theorem, which is formulated as an argument about symmetry. He notes that the proof does not require Zermeo's axiom (well-ordering principle). Sierpinski proves his result for any function $f(x)$ satisfying the properties
$f(x) = f(y), x - y \in \mathbb{Q}, f(x) \neq f(y), x - y \notin \mathbb{Q}$.
Here is a fuller rendering of Sierpinski's argument:
Suppose $f : [\mathbb{R}]^{\leq \omega} \rightarrow \mathbb{R}$. Let $E_x = \lbrace x+r : r \in \mathbb{Q} \rbrace$. So $E_x = E_y \Leftrightarrow x - y \in \mathbb{Q}$. For $x \in \mathbb{R}$, let $\varphi(x)= f(E_x)$. Note $\varphi(x) = \varphi(x) \Leftrightarrow x - y \in \mathbb{Q}$.
Claim: $\varphi(x)$ is a non-measurable function. In fact, any such function $\phi(x)$ satisfying $\phi(x) = \phi(x) \Leftrightarrow x - y \in \mathbb{Q}$ is non-measurable.
Proof: Suppose $\varphi(x)$ is measurable. Then $\psi(x) = \varphi(x) - \varphi(-x)$ is also measurable, and the set $N = \lbrace x : \psi(x) > 0 \rbrace$ is a measurable set. Let $Q = \lbrace y \in Irr : y \notin N \rbrace$. Note that for all $r \in \mathbb{Q}$ and $x \in Irr$, of the the two numbers $x$ and $2r - x$, one belongs to $N$ and the other to $Q$ (since $\psi(2r - x) = \psi(x)$ for irrational $x$, while for rational $x, \psi(x) = 0$). So $N$ and $Q$ are symmetric images of each other, when one takes any point with rational abscissa as centre of symmetry.
Now let $(a, b)$ be any interval, and suppose $N \cap (a, b)$ is measurable. Let $(a_1, b_1)$ be an interval with rational endpoints such that $(a_1, b_1) \subseteq (a, b)$. Let $N_1 = N \cap (a_1, b_1), Q_1 = Q \cap (a_1, b_1)$. $N_1$ and $Q_1$ are measurable, being symmetric images and so have the same measure, which is half the length of the interval $(a_1, b_1)$, since the points in $(a_1, b_1)$ belonging neither to $N$ nor $Q$ are countably many.
So one can decompose $(a, b)$ into two sets which have the same measure on each rational interval. It follows easily (without using the axiom of Mr Zermelo) that we have reached a contradiction. So $\varphi$ cannot be measurable.