Prop: Empty set is a subset of every set.
Pf:. First consider the empty set. $\emptyset \subseteq \emptyset \iff \emptyset \bigcap \emptyset = \emptyset $ and $\emptyset \bigcup \emptyset = \emptyset$. $\emptyset \bigcap \emptyset =\{x\in \!\, \emptyset \ \text{and} \ x\in \!\, \emptyset \!\,\}=\emptyset$ and $\emptyset \bigcup \emptyset =\{x\in \!\, \emptyset \ \text{or} \ x\in \!\, \emptyset \!\,\}=\emptyset$. Thus, $\emptyset \subseteq \emptyset$.
Now consider a non-empty set $S$. $\emptyset \subseteq S \iff \emptyset \bigcap S = \emptyset $ and $\emptyset \bigcup S = S$. $\emptyset \bigcap S = \{x \in S \ \text{and} \ x \in \emptyset \} = \emptyset$ and $\emptyset \bigcup S =\{x\in \!\, S \ \text{or} \ x\in \!\, \varnothing \!\,\}=S$.
Hence, $\emptyset \subseteq S$ as required $\square$. Does my proof look right?
