I fill the gaps in the proof sketched by Noah Schweber that every uncountable Borel set of $\mathbb{R}^n$ has an uncountable null Borel subset (since the proof is too long I can't post it as a comment). As Noah noted, since very analytic set has the perfect set property, we can assume that $A$ is a perfect set.
More generally, I will prove that if $A$ is a closed and uncountable subset of $\mathbb{R}^n$, then we can find $X \subseteq A$, with $X$ perfect set of zero Lebesgue measure.
I call a cell a set which is defined, for some $a_i \leq b_i$ for $i=1, \dots, n$, as consisting of all $x=(x_1, \dots, x_n) \in \mathbb{R}^n$, with $a_i \leq x_i \leq b_i$ for all $i=1, \dots, n$.
First, we claim there exists a sequence $(S_i)$, where each $S_i$ is a finite family of pairwise disjoint cells such that:
each $S_k$ contains $2^{k-1}$ cells;
each $I \in S_k$ contains exactly two elements $J_0, J_1$ of $S_{k+1}$;
every $I \in S_k$ has sides whose length is at most $2^{-2^k}$;
for each $I \in S_k$, the set $I \cap A$ is uncountable.
To prove this, for every $v=(v_1, \dots, v_n)$, with $v_1, \dots, v_n\in \mathbb{Z}$, and every nonnegative integer m, let $I_{m,v}$ the dyadic cell consisting of all $x=(x_1, \dots, x_n) \in \mathbb{R}^n$ such that for all $i=1, \dots, n$ we have
\begin{equation}
2^{-m}v_i \leq x_i \leq 2^{-m}(v_i + 1).
\end{equation}
Assume that for some $m$ and $v$, the set $I_{m,v} \cap A$ is uncountable. Then we claim that there are infinitely many integers $p > m$ such that, for each such $p$ there are $s, t \in \mathbb{Z}^n$, such that $I_{p,s}$ and $I_{p,t}$ are disjoint, both contained in $I_{m,v}$, and such that both $I_{p,s} \cap A$ and $I_{p,t} \cap A$ are uncountable. If not, there would exist some integer $N$ such that, for all $p > N$, there exists a dyadic cell $C_p$ of side of length $3 \times2^{-p}$ which contains all the cells of side $2^{-p}$ which have an uncountable intersection with $A$. We would get a sequence of compact sets $(C_p)_{p > N}$, with $C_{p+1} \subset C_p$, so that $\bigcap_{p>N} C_p$ contains exactly one point. But then $A$ would be countable, a contradiction.
From the property just proved, we can easily construct our sequence $(S_i)$.
Now, define $X= \bigcap_{i \in \mathbb{N}} (\bigcup S_i)$. Clearly $X$ is closed, since each $\bigcup S_i$ is. Moreover, if $x \in X$, then for each $k$, $x$ belongs to some $I \in S_k$. Since each point $y \in S_k$ has a distance from $x$ which is not greater than $n \times 2^{-2^k}$, $x$ turns out to be a limit point of a sequence of points of $A$, and since $A$ is closed, we conclude that $x \in A$. Finally, let $J_0, J_1$ be the elements of $S_{k+1}$ contained in $I$, and $x \in J_0$. Then there is some $y \in J_1 \cap X$ (since a decreasing sequence of compact sets has nonempty intersection). The distance between $x$ and $y$ is not greater than $n \times 2^{-2^{k}}$. We conclude that $x$ is a limit point of $X$, so that $X$ is perfect. To see that $X$ has zero Lebesgue measure just note that each $S_k$ has measure at most $2^{k-1} \times 2^{-n 2^k}$. QED
