Proof
Assume sets A and B satisfy \(A \subseteq B\text{.}\) Then, notice that
\begin{equation*}
A \cap (B-A) = \emptyset
\end{equation*}
and
\begin{equation*}
B = A \cup (B-A).
\end{equation*}
Therefore, by subadditivity and nonnegativity 4.3.2
\begin{gather*}
0 \le P(B-A)\\
P(A) \le P(A) + P(B-A) \\
P(A) \le P(B)
\end{gather*}