Solution: Take $X = \mathbb{R} ^2$ with the Euclidean topology. Let $B[\mathbf{x} ,r] \left( B(\mathbf{x} ,r) \right) $ denotes the closed (open) ball centered at $\mathbf{x} $ and radius $r$ Consider the set \[ A = B[(-1,0), 0.5] \cup [-0.5, 0.5] \cup B[(1,0), 0.5]. \] Then the interior of the set $A$ will be two open balls, namely \[ \text {Int}(A) = B((-1,0), 0.5) \cup B((1,0),0.5) \] which is not connected. Thus, $A$ is connected does not imply $\text{Int}(A)$ is connected.

On the other hand, if $A$ is connected, then its closure will be connected. We will prove it by two methods. The first method uses one of the characterizations of connectedness. Let $\bar{A} $ denotes the closure of $A$. Let \[ f: \bar{A} \to \left( \{ 0,1 \}, \mathcal{T} _d \right) \] be continuous. Since $f$ is continuous its restriction to any subset will be continuous and hence $f\big|_{A} : A \to \left( \{ 0,1\}, \mathcal{T} _d \right) $ is continuous. Since $A$ is connected, $f$ must be constant. Without loss of generality, we assume that $f(A) = \{ 0 \} $. Now we claim that $f\left( \bar{A} \right) = \{ 0 \} $. Since $f$ is continuous, we have \[ f\left( \bar{A} \right) \subset \overline{f(A)} = \{ 0 \}. \] Therefore, $f\left( \bar{A} \right) = \{ 0 \} $ and hence, $\bar{A} $ is connected.

Secondly, we can use the definition of connectedness. Suppose that $\bar{A} = E \cup F$, where $\bar{E} \cap F = E \cap \bar{F} = \emptyset $. Suppose that $E\neq \emptyset $. We will prove that $F = \emptyset $. Note that we can write \[ A = \left( A \cap E \right) \cup \left( A \cap F \right). \] Since $ E \neq \emptyset $ so $A \cap E \neq \emptyset $. As $A$ is connected, $A \cap F$ must be an emptyset. Now, $A \subseteq E \cup F$ and $A \cap F = \emptyset $, which implies \[ A \subseteq E \implies \bar{A} \subseteq \bar{E}. \] Now note that \[ \bar{A} = E \cup F \implies F = F \cap \bar{A} \subseteq F \cap \bar{E} = \emptyset . \] Thus, $F= \emptyset $ and hence, $\bar{A} $ is connected.