I have recently been introduced to the concept of convergence in probability. If $\{X_n\}$ and $\{Y_n\}$ converge in probability to $X$ and $Y$ respectively, then I would like to prove the following properties:
$$X_n+Y_n \overset{p}{\to} X+Y \ \ \text{ and } \ \ X_nY_n \overset{p}{\to} XY$$
I know that given a continuous function $g(\vec{x})$ of a vector argument, and a sequence of vectors that converges in probability, $\vec Z_n \overset{p}{\to} \vec Z$ , we have: $$ g(\vec Z_n)\overset{p}{\to} g(\vec Z) $$
Because multiplication and addition are continuous mappings, it is clear that if ${X_n\choose Y_n} \overset{p}{\to}{ X\choose Y} $ it would follow that $X_n+Y_n \overset{p}{\to} X+Y$ and $X_nY_n \overset{p}{\to} XY$.
However, I have not been able to prove this property from the convergence of $X_n$ and $Y_n$.
Given $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$, how does one prove that ${X_n\choose Y_n} \overset{p}{\to}{ X\choose Y} $?
While trying to find a solution, I came across this post from $9$ years ago. There is a key step in the logic that I do not understand, (a similar step is seen in this post).
$$P\big(|(X_n,Y_n)-(X,Y)|>\epsilon\big)\le P\big(|X_n-X|>\epsilon/\sqrt{2}\big)+P\big(|Y_n-Y|>\epsilon/\sqrt{2}\big)$$
I can easily show the desired result using this inequality but I cannot prove it myself. Where does this inequality come from?
