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