Given $X \subset \mathbb{R}^{n}$ connected, if $A \subset X$ is such that $\partial A \cap X = \emptyset$, then $A = \emptyset$ or $A=X$ 
Consider the statements:
(i) $X \subset \mathbb{R}^{n}$ is connected;
(ii) If $A \subset X$ is such that $\partial A \cap X = \emptyset$, then $A = \emptyset$ or $A=X$.
Show that (i) implies (ii), but the converse is false.

Suppose $A \subset X$ is such that $A \neq \emptyset$ and $A \neq X$. Since $A \subset X$, $A \cup \partial A = \overline{A} \subset \overline{X}$. Thus, $(\partial A \cap \overline{X}) \neq \emptyset$, but $\overline{X} = X \cup X'$, so $(\partial A \cap X) \cup (\partial A \cap X') \neq \emptyset$.
That's all I got. My idea is to cover $X \subset A_{1} \cup A_{2}$ so that $(\partial A \cap X)$ would be part of the set $A_{1}$ in order to use the connectedness. Any hint?
For the converse, I have no idea. I'm trying to get counterexamples in $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$. I'm thinking first of a set that satisfies (ii), but in all cases I got a connected set.
 A: Assume that $X$ is connected and let $A \subseteq X$ such that $\partial A \cap X = \emptyset$. We will show that $A$ is both open and closed in $X$.
Recall that $\overline{A} = \operatorname{Int} A \cup \partial A$.
We have $A = X \cap \operatorname{Int} A$. Indeed, clearly $X \cap \operatorname{Int} A \subseteq X \cap A = A$. Conversely, $$A = X \cap A \subseteq X \cap (\operatorname{Int} A \cup \partial A) = (X \cap \partial A) \cup (X \cap \operatorname{Int} A) = X \cap \operatorname{Int} A$$
so we get that $A$ is open in $X$.
Also we have $A = X \cap \overline{A}$ because
$$X \cap \overline{A} = X \cap (\operatorname{Int} A \cup \partial A) = (X \cap \partial A) \cup (X \cap \operatorname{Int} A) = A$$
so $A$ is closed in $X$.
If $A \ne \emptyset, X$ then $(A, X\setminus A)$ would be a separation of $X$, which is impossible since $X$ is connected.

For a counterexample to the converse let $X$ be any disconnected closed subset of $\mathbb{R}^n$. If $A \subseteq X$ is such that $\partial A \cap X = \emptyset$, we have $\partial A \subseteq \overline{A} \subseteq \overline{X} = X$ so $\partial A = \partial A \cap X = \emptyset$.
It follows that $A$ is both open and closed in $\mathbb{R}^n$. Indeed, we have $\overline{A} = \operatorname{Int} A \cup \partial A = \operatorname{Int} A$ so $\operatorname{Int} A  = A = \overline{A}$.
Since $\mathbb{R}^n$ is connected, we conclude $A = \emptyset$ or $A = \mathbb{R}^n$, but the latter is impossible because $A \subseteq X \ne \mathbb{R}^n$.
Hence $A = \emptyset$ so $X$ satisfies the property $(2)$ but it isn't connected.
