Solution: We need to show that \[ \bigcup_{A\in \mathcal{A} } \text{Int}(A)\subseteq\text{Int}\left( \bigcup_{A\in \mathcal{A}} A \right). \] Let $x\in \bigcup_{A\in \mathcal{A} } \text{Int}(A)$. This means there exists $A\in \mathcal{A} $ such that $x\in \text{Int }(A)$ which implies there exists an open neighborhood $U$ of $x$ such that $U\subseteq A$. So, \[ x\in U \subseteq A \subseteq \bigcup_{A\in \mathcal{A} } A. \] Thus, $x\in \text{Int}\left( \bigcup_{A\in \mathcal{A}} A \right)$.

Note that the reverse inequality need not be true. For example, take \[ A = [0,1],\text{ and } B = [1,2]. \] Then \[ \text{Int}(A) = (0,1),~\text{Int}(B) = (1,2). \] But, \[ \text{Int}(A\cup B) = (0,2) \text{ and } \text{Int}(A) \cup \text{Int}(B) = (0,2)\setminus \{1\}. \]