Is it true that if the property "being a closure of its interior" holds for a set, then it also holds for each connected component of that set? Suppose that $A\subseteq\mathbb{R}^k$ is the closure of its interior. Is it true, then, that each connected component of $A$ is the closure of its respective interior? I.e., does it hold that
$$\tag{1} A = \mathrm{clos}(\mathrm{int}(A)) \qquad \Longrightarrow \qquad \left(\forall\, \text{connected component $C$ of $A$} \ : \ C = \mathrm{clos}(\mathrm{int}(C))\right)\quad ?$$
 A: No, it is not true. For exemple, consider $$A = \lbrace 0 \rbrace \cup \bigcup_{n \geq 1} \left[ \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right] \subset \mathbb{R} \, \text{.}$$ Then we have $$\mathring{A} = \bigcup_{n \geq 1} \left( \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right) \, \text{,}$$ and hence $\overline{\mathring{A}} = A$. However, we have $$\overline{\mathring{\lbrace 0 \rbrace}} = \varnothing \neq \lbrace 0 \rbrace \, \text{.}$$
Edit: I have been asked in the comments if the property is true if one also assumes that each connected compact has nonempty interior. The answer is still no. For example, consider $$A = C \cup \bigcup_{n \geq 1} \left( \left[ \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right] \times [-1, 1] \right) \subset \mathbb{R}^{2} \, \text{,}$$ where $$C = \left( [-1, 0) \times [-1, 0] \right) \cup \left( \lbrace 0 \rbrace \times [-1, 1] \right) \, \text{.}$$ We have $\overline{\mathring{A}} = A$. Moreover, the connected components of $A$ are precisely $C$ and the sets $$\left[ \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right] \times [-1, 1] \, \text{,}$$ with $n \geq 1$, which all have nonempty interior. However, we have $$\overline{\mathring{C}} = [-1, 0]^{2} \neq C \, \text{.}$$
