Let $A \subset \mathbb{R}^d$ ($d \geq 2$) be an uncountable, closed set, such that every point of $A$ is an accumulation point of $A$ ($A$ is perfect).
I want to know whether there is some $x_0 \in A$ and some direction $v$ such that $A$ accumulates at $x_0$ "from both sides" of the hyperplane perpendicular to $v$.
Precisely, for $v \in \mathbb{R}^d$, let us write $$ H_v := \{ x \in \mathbb{R}^d \colon \langle x, v\rangle > 0\}, $$ noting the strict inequality.
I would like to know whether there necessarily exists some $x_0 \in A$ and some $v \neq 0$ such that $$ x_0 \in \overline{A \cap (x_0 + H_v)} \cap \overline{A \cap (x_0 + H_{-v})}, $$ where the overline denotes the closure and $c + B = \{ c + b \colon b \in B\}$; that is, $x_0 + H_v$ is the half-plane $H_v$ "attached" to the base point $x_0$.
If we would just require that there is at least one point of accumulation (no perfectness, no uncountability), then the claim is false. The idea here is that if we consider the set $A := \{0\}\cup \{ (n^{-1} , n^{-1/2}) \colon n \in \mathbb{N} \}$, then $0$ is a point of accumulation, but for any $v \neq 0$, one can show that $0$ is either not a point of accumulation of $A \cap H_v$, or not a point of accumulation of $A \cap H_{-v}$.
By iterating this construction, I think one can also construct a (non-closed) countable set such that every point is an accumulation point, but such that my desired property does not hold.
I have solved my own question in the affirmative. Actually, the following stronger claim is true:$\newcommand{\R}{\mathbb{R}}\newcommand{\N}{\mathbb{N}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\diam}{\operatorname{diam}}$
Proposition: Let $d \in \N$, and let $A \subset [0,1]^d$ be uncountable and closed. Then there is some $x_0 \in A$ and some $i \in \{ 1,\dots,d \}$, such that $x_0$ is an accumulation point of both $A \cap (x_0 + H_i^{(d,+)})$ and $A \cap (x_0 + H_i^{(d,-)})$, where $$ H_i^{(d,+)} := \{ x \in \R^d \colon x_i > 0 \} \qquad \text{and} \qquad H_i^{(d,-)} := \{ x \in \R^d \colon x_i < 0 \} . $$
Proof: We prove the claim by induction on $d \in \N$.
Base case ($d = 1$): For the base case, we will not use that $A$ is closed; this will be used in the induction step below.
Let us assume towards a contradiction that the claim is false; this implies that $A = A_1 \cup A_2$, where $$ A_1 := \{ x \in A \colon \exists \, \delta > 0 : (x, x + \delta] \cap A = \emptyset \} %\\ \quad \text{and} \quad A_2 := \{ x \in A \colon \exists \, \delta > 0 : [x - \delta, x) \cap A = \emptyset \} . $$ Thus, $A_1$ or $A_2$ is uncountable. We only consider the case that $A_1$ is uncountable; the other case can be handled in a similar way. For each $x \in A_1$, we can choose some $q_x \in \Q_{> x}$ such that $(x, q_x] \cap A = \emptyset$. Since $A_1$ is uncountable while $\Q$ is not, there must be some $q \in \Q$ such that $A_1 ' := \{ x \in A_1 \colon q_x = q \}$ is uncountable. Note that $x < q_x = q$ for all $x \in A_1 '$, so that $u := \sup A_1 ' \leq q$ is well-defined.
Since $A_1 '$ is infinite, there is some $x \in A_1 '$ satisfying $x < u$. By definition of the supremum, there is thus some $y \in A_1 ' \subset A$ with $x < y \leq u \leq q$. However, by choice of $q_x$ and since $q_x = q \geq u$, we have $\emptyset = A \cap (x, q_x] \supset A \cap (x, u] \ni y$, which is the desired contradiction.
Induction step ($d \to d+1$): Suppose that the claim holds for some $d \in \N$; we prove that it also holds for $d + 1$. For each $x \in [0, 1]$, set $$ A_x := \big\{ y \in [0, 1]^{d} \colon (x, y) \in A \big\}, $$ and note that $A_x \subset [0, 1]^d$ is closed. We first consider the case where $A_x$ is uncountable for some ${x \in [0, 1]}$. By the induction hypothesis, we see that there exist $y \in A_x$ and $i \in \{ 1, \dots, d \}$ such that $y$ is an accumulation point of $B_1 := A_x \cap (y + H_i^{(d,+)})$ and of $B_2 := A_x \cap (y + H_i^{(d,-)})$. Thus, there are sequences $(y^{(n)})_{n \in \N} \subset B_1$ and $(z^{(n)})_{n \in \N} \subset B_2$ satisfying $y = \lim_n y^{(n)} = \lim_n z^{(n)}$. Therefore, ${(x, y) = \lim_n (x, y^{(n)}) = \lim_n (x, z^{(n)})}$, and furthermore $(x, y^{(n)}) \in A \cap \big( (x,y) + H_{i+1}^{(d+1,+)} \big)$ and ${(x, z^{(n)}) \in A \cap \big( (x,y) + H_{i+1}^{(d+1,-)} \big)}$, proving the claim of the proposition.
It remains to consider the case where $A_x$ is countable for each $x \in [0, 1]$. For the projection $$ \pi : A \subset [0, 1]^{d+1} \cong [0, 1] \times [0, 1]^d \to [0, 1], (x,y) \mapsto x, $$ this implies that $\pi^{-1}(\{ x \}) = \{ x \} \times A_x$ is countable for each $x \in [0, 1]$. Since $A = \bigcup_{x \in \pi(A)} \pi^{-1}(\{ x \})$, this implies that $B := \pi(A) \subset [0, 1]$ is uncountable. We claim that in this case, the claim of the proposition holds for $i = 1$; assume towards a contradiction that this is \emph{not} true. Then, for each $x \in B$, the following holds: $$ \forall \, y \in [0, 1]^d \,\, \exists \, \delta_{x,y} > 0 \text{ and } \sigma_{x,y} \in \{ \pm 1 \} \,\, : \,\, %\text{ such that } A \cap \big( [x + \sigma_{x,y} \, (0, \delta_{x,y})] \times [y + (-\delta_{x,y}, \delta_{x,y})] \big) = \emptyset . \tag{$\ast$} $$ Indeed, if $(x,y) \notin A$, this holds since $A$ is closed; if otherwise $(x,y) \in A$, then this follows since we assume that the claim of the proposition does not hold for $i = 1$.
For brevity, set $J_{x,y} := y + (- \delta_{x,y}, \delta_{x,y})$, and note for each fixed $x \in B$ that $(J_{x,y})_{y \in [0,1]^d}$ is an open cover of the compact set $[0, 1]^d$. Thus, the Lebesgue number lemma (see https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma) shows that there is some $N_x \in \N$ such that for every set $\Lambda \subset [0,1]^d$ with $\diam \Lambda \leq N_x^{-1}$, we have $\Lambda \subset y + (- \delta_{x,y}, \delta_{x,y})$ for some $y \in [0,1]^d$. Here, the diameter of $\Lambda$ is calculated using the $\ell^\infty$-norm; that is, $\diam \Lambda := \sup_{\lambda,\mu \in \Lambda} \| \lambda - \mu \|_{\ell^\infty}$. Since $B$ is uncountable, there is some $N \in \N$ such that the set $B_1 := \{ x \in B \colon N_x = N \}$ is uncountable as well.
Set $I := \{ 0, \dots, N - 1 \}^d$ and $Q_i := i + [0, N^{-1}]^d$ for $i \in I$, noting that $[0, 1]^d = \bigcup_{i \in I} ( i + [0,N^{-1}]^d )$ and $\diam Q_i \leq N^{-1}$. Thus, for all $x \in B_1$ and $i \in I$, there is some $y(x,i) \in [0,1]$ such that ${Q_i \subset J_{x, y(x,i)}}$. Choose some $\varrho_x \in \Q$ satisfying $0 < \varrho_x < \min_{i \in I} \delta_{x, y(x,i)}$, and define ${\sigma_x := (\sigma_{x, y(x,i)})_{i \in I} \in \{ \pm 1 \}^I}$. Since $(\varrho_x, \sigma_x)$ belongs to the countable set $\Q_{> 0} \times \{ \pm 1\}^I$ and since $B_1$ is uncountable, there are $\varrho \in \Q_{> 0}$ and $\sigma = (\sigma_i)_{i \in I} \in \{ \pm 1 \}^I$ such that the set $$ B_2 := \{ x \in B_1 \colon \varrho_x = \varrho \text{ and } \sigma_x = \sigma \} $$ is uncountable. As a special case of the base case of the induction (which holds for general uncountable, not necessarily closed sets), we see that there is some $x_0 \in B_2$ and a sequence $(x_n)_{n \in \N} \subset B_2 \setminus \{ x_0 \}$ such that $x_n \to x_0$. Because of $x_n \in B_2 \subset B_1 \subset B = \pi(A)$, we can choose for each $n \in \N$ some $y_n \in [0, 1]^d$ satisfying $(x_n, y_n) \in A$. By passing to a subsequence, we can assume that $y_n \to y_0$ for some $y_0 \in [0, 1]^d$. Since $(x_n, y_n) \in A$ and since $A$ is closed, this implies $(x_0, y_0) \in A$ as well.
Next, since $y_n \in [0, 1]^d = \bigcup_{i \in I} Q_i$ and since $I$ is finite, we can assume (by passing to a further subsequence) that there is some $i \in I$ with $y_n \in Q_i$ for all $n \in \N$. Since $Q_i$ is closed, this implies that ${y_0 \in Q_i}$ as well. Now, note because of $x_n \in B_2 \subset B_1$ that $Q_i \subset J_{x_n, y(x_n, i)}$ and $\varrho = \varrho_{x_n} < \delta_{x_n, y(x_n, i)}$, and finally ${\sigma_{x_n, y(x_n, i)} = (\sigma_{x_n})_i = \sigma_i}$ for all $n \in \N_0$. Therefore, Equation $(\ast)$ from above shows that $$ \emptyset = A \cap \big( [x_n + \sigma_{x_n, y(x_n, i)} \, (0, \delta_{x_n, y(x_n, i)})] \times J_{x_n, y(x_n, i)} \big) \supset A \cap \big( [x_n + \sigma_i \, (0, \varrho)] \times Q_i \big) \qquad \forall \, n \in \N_0 . $$ Because of $y_n \in Q_i$ and $(x_n, y_n) \in A$ for all $n \in \N_0$, we thus see that $$ x_0 \notin x_n + \sigma_i \, (0, \varrho) \quad \text{and} \quad x_n \notin x_0 + \sigma_i \, (0, \varrho) \quad \forall \, n \in \N . $$ Hence, $x_0 - x_n, x_n - x_0 \notin \sigma_i \, (0, \varrho)$, which implies because of $x_n \neq x_0$ and $\sigma_i \in \{ \pm 1 \}$ that $|x_n - x_0| \geq \varrho > 0$ for all $n \in \N$, in contradiction to $x_n \to x_0$. This is the desired contradiction, completing the proof.