The following is from an old homework assignment. I've left the question and answer format intact, since it does not address only OP's question.
Construct a closed set $\hat{\mathcal{C}}$ so that at the $k$th stage of the construction one removes $2^{k-1}$ centrally situated open intervals each of length $l_k$ with
$$l_1 + 2l_2 + \cdots + 2^{k-1}l_k < 1$$
a) If $l_j$ are chosen small enough, then
$$
\sum_{k=1}^\infty 2^{k-1}l_k < 1
$$
In this case, show that $m(\hat{\mathcal{C}}) > 0$, and in fact,
$$m(\hat{\mathcal{C}}) = 1 - \displaystyle\sum_{k=1}^\infty 2^{k-1}l_k$$
Proof:
First, we claim that $\hat{\mathcal{C}}$ is measurable. Denote by $O_k$ the union of the open sets removed from $[0,1]$ at step $k$ of the construction. Since the union of an arbitrary number of open sets is open, each of the $O_k$ is open. Furthermore, $O = \displaystyle\bigcup_{k=1}^\infty O_k$ is open. Now, we have that $\hat{\mathcal{C}} = [0,1] \setminus O$ is closed and therefore measurable (as all closed sets are measurable).
Now, to determine $m(\hat{\mathcal{C}})$, observe that both $O$ and $\hat{\mathcal{C}}$ are measurable ($O$ is measurable because it is open) and disjoint and that $O \cup \hat{\mathcal{C}} = [0,1]$. Then
\begin{align*}
m([0,1]) &= m(O) + m(\hat{\mathcal{C}})\\
m(\hat{\mathcal{C}}) &= m([0,1]) - m(O)\\
\end{align*}
Furthermore observe that all of the $O_k$ are open (and so measurable) and disjoint with $O = \displaystyle\bigcup_{k=1}^\infty O_k$. If we further break the $O_k$ into their constituent open subsets, these properties still hold. Hence
\begin{align*}
m(\hat{\mathcal{C}}) =& m([0,1]) - m(O)\\
=& m([0,1]) - \displaystyle\sum_{k=1}^\infty m(O_k)\\
=& 1 - \displaystyle\sum_{k=1}^\infty 2^{k-1}l_k
\end{align*}
b) Show that if $x \in \hat{\mathcal{C}}$, then there exists a sequence of points $\{x_n\}_{n=1}^\infty$ such that $x_n \notin \hat{\mathcal{C}}$, yet $x_n \rightarrow x$ and $x_n \in I_n$, where $I_n$ is a sub-interval in the complement of $\hat{\mathcal{C}}$ with $|I_n| \rightarrow 0$.
Proof: Observe first that since $\displaystyle\sum_{k=1}^\infty 2^{k-1}l_k < 1$, the tail of the series must go to zero. That is, for any $\epsilon > 0$, there exists $N$ such that $l_n < \epsilon$ for all $n \geq N$. Now, let $x \in \hat{\mathcal{C}}$. Let $\hat{\mathcal{C}}_k$ denote the $k$ stage of the construction. For each $k$, $x$ belongs to some closed subset $S_k$ of $C_k$. Let $I_k$ be the open interval removed from $S_k$ to proceed to the next step of the construction. We take any $x_k \in I_k$ to form our sequence $\{x_n\}_{n=1}^\infty$. Clearly, each $x_k$ belongs to an sub-interval in the complement of $\hat{\mathcal{C}}$. Furthermore, $|I_k| = l_k \rightarrow 0$. It remains to show that $x_n \rightarrow x$.
From the construction of $\hat{\mathcal{C}}_k$ and our selection of $x_n$, it is clear that
$$|x - x_n| < |I_n| + |S_n|.$$
By our previous observation, we know that $|I_n| = l_n \rightarrow 0$. Now
\begin{align*}
|S_n| &= \frac{1 - \displaystyle\sum_{k=1}^n 2^{k-1}l_k}{2^n}\\
&\leq \frac{1}{2^n}\\
&\rightarrow 0 \text{ as } n \rightarrow \infty
\end{align*}
Hence, $|x - x_n| \rightarrow 0$. That is, $\{x_n\}_{n=1}^\infty$ converges to $x$.
c) Prove as a consequence that $\hat{\mathcal{C}}$ is perfect and contains no open interval.
Proof: To see that $\hat{\mathcal{C}}$ is perfect, let $\epsilon > 0$ be given and consider $B(x,\epsilon)$ for any $x \in \hat{\mathcal{C}}$. We can find $N \in \mathbb{N}$ such that $S_N \subset B(x,\epsilon)$. Now, this interval must have two endpoints $a_N$ and $b_N$ (one of which could possibly be equal to $x$). By the construction of $\hat{\mathcal{C}}$, we know that the endpoints of any interval are never removed, and so $a_N, b_N \in \hat{\mathcal{C}}$. Furthermore, we have that $a_N, b_N \in S_N \subset B(x,\epsilon)$. Therefore, $x$ is not isolated.
Suppose, to the contrary, that there exists an open interval $O \in \hat{\mathcal{C}}$. Then, for any $x \in O$, there exists $\epsilon_0$ such that $B(x,\epsilon_0) \subseteq O$. Let $\epsilon < \epsilon_0$. Then, there can be no sequence $\{x_n\}_{n=1}^\infty$ of the type described in part (b) whose limit is $x$, since $B(x,\epsilon_0) \subseteq \hat{\mathcal{C}}$ implies that $|x - x_n| > \epsilon_0 > \epsilon$ for all $n$. This contradicts the conclusion of part (b), and so it must be that $\hat{\mathcal{C}}$ contains no open interval.
d) Show also that $\hat{\mathcal{C}}$ is uncountable. (Just for fun)
Proof: We claim that $\hat{\mathcal{C}}$ is in one-to-one correspondence with infinite ternary strings containing only 0s and 2s, and so is uncountable.
($\Rightarrow$) Let $x \in \hat{\mathcal{C}}$. We build a ternary string for $x$ of the desired form as follows. Consider $\hat{\mathcal{C}}_1$. When we remove the centrally situated open interval, it must be that $x$ belongs to either the left closed subinterval (in which case let first digit of the ternary string for $x$ be 0) or the right closed subinterval (in which case let first digit of the ternary string for $x$ be 2). Next, consider $\hat{\mathcal{C}}_2$. The interval of $\hat{\mathcal{C}}_1$ to which $x$ currently belongs will be divided into three subintervals, and so we append a 0 to the ternary string for $x$ if it belongs to the leftmost subinterval or a 2 if it belongs to the rightmost subinterval. Continuing in this way, we see that $x$ has an associated ternary string containing only the digits 0 and 2.
($\Leftarrow$) Let $s$ be an infinite ternary string containing only 0s and 2s. We associate can with $s$ an $x \in \hat{\mathcal{C}}$ as follows. If the first digit of $s$ is 0, we choose the left subinterval of $\hat{\mathcal{C}}_1$. If the first digit of $s$ is 2, we choose the rightmost subinterval of $\hat{\mathcal{C}}_1$. When we form $\hat{\mathcal{C}}_2$, the interval we have just chosen will be subdivided into three subintervals. If the second digit of $s$ is 0, we select the leftmost subinterval. If the second digit of $s$ is 2, we select the rightmost subinterval. Continue in this way. Since each $x \in \hat{\mathcal{C}}$ belongs to a singleton set, we see that $s$ will specify some $x \in \hat{\mathcal{C}}$.
Best Answer
We can find an example of a set which satisfies 1., 2. and 3., for example $$A:=(\Bbb Q^c\cap ([0,1/4)\cup[3/4,1)))\cup (\Bbb Q\cap [1/4,3/4)).$$
However, there is no set $A$ which satisfies the latest condition in the OP. Indeed, fix $x\in A$ and $r>0$. We can approximate the characteristic function of $B(x,r)$ pointwise by continuous functions. By dominated convergence theorem, we would have that $2\lambda(A\cap B(x,r))=\lambda(B(x,r))$, and we conclude by a similar argument as here.