Answer
$For\,\,all\,\,sets\,\,A,\,B,\,and\,\,C\,\\
if\,\,A^{c} \subseteq B\,\,then\,\,A \cup B = U \\
this\,\,is\,\,true:\\
A \cup B = U\\
talking\,\,complement\,\,we\,get\,:
(A \cup B)^c = U^c \\
if\,\,we\,\,prove\,\,this\,(A \cup B)^c = U^c\,to\,be\,true\,\,\\then\,A \cup B = U\,\,is\,true \\
(A \cup B)^c = U^c \\
L.H.S:\\
(A \cup B)^c=A^{c}\cap B^c (by\,De\,Morgan\,law)\\
\because A^c\subseteq B \Rightarrow\therefore A^{c}\cap B^c=B\cap B^c=\varnothing \\
L.H.S=\varnothing \\
R.H.S:\\
U^c=\varnothing (by\,def\,of\,universal\,set)\\
L.H.S=R.H.S=\varnothing\\
\because (A \cup B)^c = U^c \\
\therefore (A \cup B) = U (if\,A^c\subseteq B)
$
Work Step by Step
$For\,\,all\,\,sets\,\,A,\,B,\,and\,\,C\,\\
if\,\,A^{c} \subseteq B\,\,then\,\,A \cup B = U \\
this\,\,is\,\,true:\\
A \cup B = U\\
talking\,\,complement\,\,we\,get\,:
(A \cup B)^c = U^c \\
if\,\,we\,\,prove\,\,this\,(A \cup B)^c = U^c\,to\,be\,true\,\,\\then\,A \cup B = U\,\,is\,true \\
(A \cup B)^c = U^c \\
L.H.S:\\
(A \cup B)^c=A^{c}\cap B^c (by\,De\,Morgan\,law)\\
\because A^c\subseteq B \Rightarrow\therefore A^{c}\cap B^c=B\cap B^c=\varnothing \\
L.H.S=\varnothing \\
R.H.S:\\
U^c=\varnothing (by\,def\,of\,universal\,set)\\
L.H.S=R.H.S=\varnothing\\
\because (A \cup B)^c = U^c \\
\therefore (A \cup B) = U (if\,A^c\subseteq B)
$