I can only show $|X_{n}-X||Y_{n}-Y|\xrightarrow{P} 0$ for $0<\delta<1$

Question

Say $X_{n}\xrightarrow{P} X$ and $Y_{n}\xrightarrow{P} Y$ (*)

I want to show that:

$|X_{n}-X||Y_{n}-Y|\xrightarrow{P} 0$

My ideas:

Let $0<\delta\leq1$

$P( |X_{n}-X||Y_{n}-Y| \geq \delta )\leq P(|X_{n}-X|\geq\delta)+P(|Y_{n}-Y|\geq\delta)\xrightarrow{n\to \infty}0$

But I do not have any sensible ideas on how to show that this is the case for $\delta > 0$

Let me perhaps explain the background of the question:

I want to show under condition (*) that $X_{n}Y_{n}\xrightarrow{P}XY$ and as a hint, I am given that $X_{n}Y_{n}-XY=(X_{n}-X)(Y_{n}-Y)+X(Y_{n}-Y)+Y(X_{n}-X)$

So in order to show convergence in probability, I need to show it for $\delta > 0$ and not $0<\delta \leq 1$, correct?

Answer 1

If $P\{Z_n >\delta \} \to 0$ for $\delta \in (0,1)$ then, for $\delta \geq 1$ we have $P\{Z_n >\delta \} \leq P\{Z_n >\frac 1 2 \} \to 0$.

Answer 2

$$P( |X_{n}-X||Y_{n}-Y| \geq \delta )\leq P(|X_{n}-X|\geq\sqrt\delta)+P(|Y_{n}-Y|\geq\sqrt\delta)\xrightarrow{n\to \infty}0$$

The above inequality holds since $$\{|X_{n}-X||Y_{n}-Y| \geq \delta\} \subseteq \{|X_{n}-X|\geq\sqrt\delta\} \cup \{|Y_{n}-Y|\geq\sqrt\delta\}.$$