How to prove that $A\subset B\implies \bar{A}\subset \bar{B}$

How to prove that A is a subset of B implies closure of A is a subset of closure of B?

I encountered this problem when looking up the solution of Baby Rudin Chapter 2 Exercise 7:

Let $A_1,A_2,A_3,\cdots$ be subsets of a metric space.a) If $B_n=\bigcup A_i$, prove that $\bar{B_n}=\bigcup_{i=1}^n$, for $n=1,2,3,\cdots$

b) If $B=\bigcup_{i=1}^{\infty}A_i$, prove that $\bar{B}\supset \bigcup_{i=1}^{\infty}\bar{A}_i$

When I was looking up solutions of part b,It has a conclusion, It says that:

This amounts to the trivial observation that since $B\supset A_i$ for all $i$, then $\bar{A}_i\subset \bar{B} $ for all $i$, and therefore $\bar{B}\supset\bigcup_{i=1}^{\infty}\bar{A}_i$

I don't think it is that trivial. I have looked up other Q&As on MSE, but I don't think they have answered my question thoroughly(As they are all about proof verifications), so I would like to ask for a proof for this statement. Thanks for your help.