Answer
$For\,\,all\,\,sets\,\,A\,\,and\,\,B,\,\,if\,\,B\subseteq A^{c}\,\,then\,\,A\cap B=\varnothing \\
To\,\,prove\,\,that\,\,a\,\,set\,\,A\cap B\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\cap B\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A\cap B \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
x\in A\cap B\Rightarrow x\in A \,and\,x\in B\\
(by\,\,def.\,\,of\,\,inter\! section) \\
but\,B\subseteq A^{c} \\
this\,\,means\,\,if\,x\in B\Rightarrow x\in A^{c}\\
x\in A \,and\,x\in B\Rightarrow x\in A \,and\,x\in A^{c}\\
by\,\,def.\,\,of\,\,complement \\
x\in A \,and\,x\notin A \\
(this\,\,is\,\,a\,\,contradiction)\\
so\,\,A\cap B=\varnothing
$
Work Step by Step
$For\,\,all\,\,sets\,\,A\,\,and\,\,B,\,\,if\,\,B\subseteq A^{c}\,\,then\,\,A\cap B=\varnothing \\
To\,\,prove\,\,that\,\,a\,\,set\,\,A\cap B\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\cap B\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A\cap B \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
x\in A\cap B\Rightarrow x\in A \,and\,x\in B\\
(by\,\,def.\,\,of\,\,inter\! section) \\
but\,B\subseteq A^{c} \\
this\,\,means\,\,if\,x\in B\Rightarrow x\in A^{c}\\
x\in A \,and\,x\in B\Rightarrow x\in A \,and\,x\in A^{c}\\
by\,\,def.\,\,of\,\,complement \\
x\in A \,and\,x\notin A \\
(this\,\,is\,\,a\,\,contradiction)\\
so\,\,A\cap B=\varnothing
$