Let $A,B$ be in a $\sigma$-algebra $\Sigma$ of $X$. Then $B \cap A = X - ((X - A) \cup (X - B))$ by De Morgan's law and so $B \cap A$ is in $\Sigma$ too (and we see $\Sigma$ is closed under intersections). Then $B \cap (X - A\cap B) = B - A$ and so set differences are also within $\Sigma$.
Countable additivity says that if $A_1 \cap A_2 = \emptyset$, then $m(A_1) + m(A_2) = m(A_1 \cup A_2)$. $A$ and $B-A$ are disjoint. $m(A) + m(B-A) = m(A \cup (B - A)) = m(B)$ where the last step uses the fact $A \subseteq B$. Because $m(B-A) \geq 0$, we see that $m(A) \leq m(B)$.