# Convergence of two sequences in probability implies joint convergence in probability- problem with proof

Let $$X_n\xrightarrow{P} X$$ and $$Y_n\xrightarrow{P}Y$$, then we have $$(X_n,Y_n)\xrightarrow{P}(X,Y)$$. In process of proof we have the following: $$\mathbb{P}(\{||(X_n,Y_n)-(X,Y)||\geqslant\epsilon\})\leqslant\mathbb{P}(\{||X_n-X||+||Y_n-Y||\geqslant\epsilon\})\leqslant\mathbb{P}(\{||X_n-X||\geqslant\epsilon/2\})+\mathbb{P}(\{||Y_n-Y||\geqslant\epsilon/2\})$$ I don't understand the above inequalities. Could someone explain?

$$\|(X_n, Y_n)-(X,Y)\|=\|(X_n-X, Y_n-Y)\|\le \|X_n-X\|+\|Y_n-Y\|$$ is true by triangle inequality.
Hence $$\|(X_n, Y_n) - (X,Y) \| \ge \epsilon \implies \|X_n-X\|+\|Y_n-Y\| \ge \epsilon$$
Also, $$\|X_n - X\| + \|Y_n - Y\| \ge \epsilon$$ implies that $$\|X_n - X\| \ge \frac{\epsilon}2$$ or $$\|Y_n - Y\| \ge \frac{\epsilon}2$$. Supppose on the contrary that this is not true, then we have $$\|X_n - X\| < \frac{\epsilon}2$$ and $$\|Y_n - Y\| < \frac{\epsilon}2$$ and summing them up would give us a contradiction.