Showing $X_n\rightarrow_P X$ and $X_n\rightarrow_P Y$ $\implies$ $P(X=Y)=1$ A sequence of random variables $\left\{ X_n \right\}$ $\textbf{converges in probability}$ to a random variable $X$, denoted by $X_n\rightarrow_P X$, if for every $\epsilon>0$,
$$ P(|X-X_n|\geq \epsilon)\rightarrow 0 \quad \text{ as } \quad n\rightarrow \infty.  $$
I am reading the proof of the following:
Let $\left\{ X_n \right\}$ be a sequence of random variables and let $X$ and $Y$ be random variables. If $X_n\rightarrow_P X$ and $X_n\rightarrow_P Y$ then $P(X=Y)=1$.
The proof begins as follows:
Let $\epsilon >0$. By the triangle inequality we have
$$ P(|X-Y|\geq \epsilon) \leq P(|X-X_n|\geq \epsilon/2) + P(|Y-X_n|\geq \epsilon/2)   $$
***I am struggling to see that last inequality. We can choose an epsilon such that $|X-X_n|\geq \epsilon/2$ and $|Y-X_n|\geq \epsilon/2$. By the triangle inequality we have $|X-Y|\leq |X-X_n|+|Y-X_n|$, but why does this sum split in the probabilities? Does it somehow have to do with the subadditivity of the probability measure?
Thank you!

The event $\{|X-Y|\geq \varepsilon\}$ is contained in the union of the events $\{|X-X_n|\geq \frac{\varepsilon}{2}\}$ and $\{|X_n-Y|\geq \frac{\varepsilon}{2}\}$.
To show this, observe that if $|X-X_n|<\frac{\varepsilon}{2}$ and $|X_n-Y|<\frac{\varepsilon}{2}$ then
$$ |X-Y|\leq |X-X_n|+|X_n-Y|<2\cdot \frac{\varepsilon}{2}=\varepsilon $$
Hence if $|X-Y|\geq \varepsilon$ then one of $|X-X_n|,|X_n-Y|$ must be $\geq \frac{\varepsilon}{2}$, which proves the inclusion.
Then the subadditivity of probability shows that
$$ \mathbb{P}(|X-Y|\geq \varepsilon)\leq \mathbb{P}\Big(|X-X_n|\geq \frac{\varepsilon}{2}\Big)+\mathbb{P}\Big(|X_n-Y|\geq \frac{\varepsilon}{2}\Big)$$
