Solution: To prove that \( B \) is connected, we will proceed by contradiction. Assume that \( B \) is not connected. Then, there exist two non-empty disjoint open sets \( U \) and \( V \) in \( B \) such that \( B = U \cup V \).

Since \( A \subset B \), we have \( A = (A \cap U) \cup (A \cap V) \). Because \( A \) is connected, either \( A \cap U = \emptyset \) or \( A \cap V = \emptyset \). Without loss of generality, assume \( A \cap V = \emptyset \). Then, \( A \subset U \).

Now, consider the closure \( \overline{A} \). Since \( U \) is open in \( B \) and \( A \subset U \), it follows that \( \overline{A} \subset \overline{U} \). But \( B \subset \overline{A} \), so \( B \subset \overline{U} \). This implies that \( V \subset \overline{U} \).

However, \( U \) and \( V \) are disjoint open sets in \( B \), and \( V \subset \overline{U} \) implies that \( V \) must intersect \( U \), which is a contradiction. Therefore, our assumption that \( B \) is not connected must be false, and \( B \) is indeed connected.