Piggybacking on this answer, here are two more notions of convergence where finite sets can converge to "larger" sets. Let $d(x,A) = \inf_{y \in A} \|x-y\|$.
Convergence in Hausdorff metric, which for $\mathbb{R}^n$ has the representation
$$
d_H(A,B) = \sup_{x \in \mathbb{R}^n} |d(x,A) - d(x,B)|,
$$
i.e., $A_k \xrightarrow{H} A$ if $d_H(A_k, A) \to 0$. This concept of set convergence is essentially classical now, and it's used in all corners of analysis and geometry.
Convergence in the sense of Painlevé-Kuratowski. For the sequence of sets $\{A_k\}_{k=1}^{\infty}$, define
\begin{align}
\limsup_{k \to \infty} A_k &= \{x : \liminf_{k \to \infty} d(x, A_k) = 0\}, \\
\liminf_{k \to \infty} A_k &= \{x : \limsup_{k \to \infty} d(x, A_k) = 0\},
\end{align}
and say that $A_k \xrightarrow{PK} A$ if
$$
\limsup_{k \to \infty} A_k \subset A \subset \liminf_{k \to \infty} A_k.
$$
Both notions are sensitive to the topology of $\mathbb{R}^n$, which is relevant to your original question. Since $d_H(A, \mathrm{cl\ } A) = 0$, the Hausdorff metric is only a "metric" on the nonempty closed subsets of $\mathbb{R}^n$. On $\mathbb{R}^n$, Painlevé-Kuratowski convergence can be induced by a metric such as
$$
d_{PK}(A,B) =
\int_0^{\infty}
\sup_{\|x\| \leq r} |d(x,A) - d(x,B)| \ e^{-r} \ dr,
$$
and one can verify that $d_{PK} \leq d_H$. Hence, $A_k \xrightarrow{H} A$ implies $A_k \xrightarrow{PK} A$, although the reverse is not true.
In broad terms, the difference in how the Hausdorff and Painlevé-Kuratowski convergences topologize the closed subsets of $\mathbb{R}^n$ is analogous to the difference between "uniform convergence" and "uniform convergence on compact sets" of functions. The following example is illustrative.
Example. Taking the closed balls $B_\rho = \{x : \|x\| \leq \rho \}$, it is natural to ask if $B_\rho \to \mathbb{R}^n$ as $\rho \to \infty$. We first observe that $d(x, \mathbb{R}^n) = 0$ for all $x$. On the other hand, $d(x, B_\rho) = \max\{0, \|x\|-\rho\}$, so by making $\|x\|$ arbitrarily large we see that $d_H(B_\rho, \mathbb{R}^n) = +\infty$ for each $\rho$. Conversely, $\sup_{\|x\| \leq r} d(x, B_\rho) = \max\{0, r-\rho\}$, which implies that
$$
d_{PK}(B_\rho, \mathbb{R}^n) = \int_\rho^\infty (r - \rho) e^{-r} \ dr = e^{-\rho},
$$
so $d_{PK}(B_\rho, \mathbb{R}^n) \to 0$ as $\rho \to \infty$.
In other words, compact sets cannot converge to $\mathbb{R}^n$ under Hausdorff metric, but they can in the sense of Painlevé-Kuratowski.
In your example, taking $A_k = \{j \cdot 2^{-k} : 1 \leq j < 2^k \}$ and $A = [0,1]$, we obtain $d_H(A_k, A) = 2^{-(k+1)}$, which is achieved at any "midpoint" between the dyadic rationals of $A_k$. Hence, $A_k \xrightarrow{H} A$ and consequently $A_k \xrightarrow{PK} A$.
Another example: let $\mathbb{Q} = \{q_1, q_2, \dots\}$ be an enumeration of the rationals, and let $A_k = \{q_1, \dots, q_k\}$. Then $d_H(A_k, \mathbb{R}) = +\infty$ for each $k$, while $d_{PK}(A_k, \mathbb{R}) \to 0$.
