Prove $A^{c}\cup B=U\implies A\subseteq B$ In the case of $\, p\in U\implies p\in A^{c}\cup B\,$, $$\begin{align} p\in A & \implies p\in A\; \land\; \color{red}{p\in U}\\\ & \implies p\in A\;\land\;\color{red}{(p\in A^{c}\;\lor\; p\in B)} \\\ & \implies (p\in A\;\land\; p\in A^{c})\;\lor\;(p\in A\;\land\; p\in B) \\\ & \implies p\in A\;\land\; p\in B \\\ & \implies p\in B.\\\ \end{align}$$ However, when it comes to $p\in A^{c}\cup B\implies p\in U$, I don't know where to start from. Can you give me some hints?

Assuming $A^{c}\cup B=U$

Let $x\in A$ we have $x\in U$ so $x \in A^{c} \cup B$ and $x\
otin A^{c}$ so $x \in B$, i.e. $A \subset B$

So we have : $A^{c}\cup B=U\implies A\subseteq B$