Inequality involved in Slutsky's Proof

Question

Why is the following inequality true? $$P(X_n+Y_n\leq x)\leq P(\{X_n+Y_n\leq x\} \ \cap \{|Y_n-c|<\epsilon\})+P(|Y_n-c|\geq\epsilon)$$ I've been trying to think of it as: $$P(A)\leq P(A\cap B)+P(B^c)$$ But there is a step in this conclusion I'm not seeing.

edit for the proof, but is there a shorter way to do this that is slightly easier to see?

Proof: $$P(A^c \cup B^c)\leq P(A^c)+P(B^c)\Rightarrow -P(A^c\cup B^c)\geq-P(A^c)-P(B^c)$$ $$\Rightarrow 1-P(A^c\cup B^c)\geq1-P(A^c)-P(B^c)$$ Thus $$P(A\cap B)=1-P((A\cap B)^c)=1-P(A^c\cup B^c)\geq 1-P(A^c)-P(B^c)$$ $$=1-(1-P(A))-P(B^c)=P(A)-P(B^c)$$ $$\Rightarrow P(A\cap B)\geq P(A)-P(B^c)$$ $$\Rightarrow P(A\cap B)+P(B^c) \geq P(A). \ \ \square$$

Answer 1

Note that $$ P(X_n+Y_n\leq x)= P(\{X_n+Y_n\leq x\} \ \cap \{|Y_n-c|<\epsilon\})+P(\{X_n+Y_n\leq x\}\cap \{|Y_n-c|\geq\epsilon\}) $$ by the law of total probability but $$ P(\{X_n+Y_n\leq x\}\cap \{|Y_n-c|\geq\epsilon\})\leq P(|Y_n-c|\geq\epsilon) $$ since one event is a subset of the other. The result follows.

Answer 2

This is a consequence of the following fact: $$ \mathbb{P}(A) = \mathbb{P}(A\cap B)+\mathbb{P}(A\cap B^c) \leq \mathbb{P}(A\cap B)+\mathbb{P}(B^c) $$ (the first equality is the law of total probability, and the inequality follows from $A\cap B^c \subseteq B^c$) applied to $A = \{X_n+Y_n \leq x\}$ and $B=\{\lvert Y_n-c\rvert < \varepsilon\}$.