The meaning of $P$ is monotone relative to set containment is this:
If $A \subseteq B$ then $P(A)\leq P(B)$.
Now here for every $\omega\in\Omega$, we have
$$\begin{align*}
A &= \left\{\omega : \left| (X_n(\omega) + Y_n(\omega)) - ( X(\omega) + Y(\omega) ) \right| \geq \epsilon \right\}, \text{ and}\\
B &= \{ \omega : \left| X_n(\omega) - X(\omega) \right| + \left| Y_n(\omega) - Y(\omega) \right| \geq \epsilon \}.
\end{align*}$$
From your first inequality and for every $\omega\in\Omega$ and $\varepsilon >0$, we have
$$\left\{ \left| (X_n(\omega) + Y_n(\omega) - (X(\omega) + Y(\omega)) \right| \geq \varepsilon \right\} \subseteq \left\{ \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| \geq \varepsilon \right\}.$$
So from the statement above
$$P\left( \left| (X_n(\omega) + Y_n(\omega) - (X(\omega) + Y(\omega)) \right| \geq \varepsilon \right) \leq P\left( \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| \geq \varepsilon \right).$$
To show the last inequality, first we prove that for every for every $\omega\in\Omega$ and $\varepsilon >0$ we have
$$\begin{multline*}
\{\omega : \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| \geq \varepsilon \}\\
\subseteq \{\omega : \left|X_n(\omega) - X(\omega) \right| \geq \varepsilon/2\} \cup \{\omega : \left|Y_n(\omega) - Y(\omega) \right| \geq \varepsilon/2\}.
\end{multline*}$$
This can be simply shown by contradiction, i.e. if
$$\{ \omega : \left|X_n(\omega) - X(\omega) \right| < \varepsilon/2 \} \text{ and } \{\omega : \left|Y_n(\omega) - Y(\omega) \right| < \varepsilon/2 \}$$
then we can sum them up to get $\{ \omega : \left|X_n(\omega) - X(\omega)\right| + \left|Y_n(\omega) - Y(\omega)\right| < \varepsilon \}$, which is a contradiction.
Hence,
$$\begin{align*}
P\bigl( \{ \omega : &\left| X_n(\omega)-X(\omega) \right| + \left| Y_n(\omega) - Y(\omega) \right|\geq \varepsilon \} \bigr)\\
&\leq P\left( \{\omega : \left| X_n(\omega) - X(\omega) \right| \geq \varepsilon/2\} \cup \{\omega : \left| Y_n(\omega) - Y(\omega) \right| \geq \varepsilon/2 \} \right)\\
&\leq P\left( \{\omega : \left| X_n(\omega) - X(\omega) \right| \geq \varepsilon/2\}\right) + P\left( \{\omega : \left| Y_n(\omega) - Y(\omega) \right| \geq \varepsilon/2 \} \right),
\end{align*}$$
where the last inequality comes from $P(C \cup D) \leq P(C) + P(D)$. Cheers!
If possible, suppose there exists a UMP test $\phi^*$ (say) of level $\alpha$ for testing $H_0:\theta=\theta_0$ vs $H_1:\theta\ne \theta_0$. Then $\phi^*$ will also be UMP level $\alpha$ for testing $H_0:\theta=\theta_0$ against $H_1':\theta>\theta_0$ as well as $H_1'':\theta<\theta_0$.
But a UMP level $\alpha$ test for $(H_0,H_1')$ is
$$
\phi_1(\mathbf X)=\begin{cases} 1 &,\text{ if }\frac{\sqrt n(\overline X-\theta_0)}{\sigma_0}>z_{\alpha}
\\ 0 &,\text{ otherwise }
\end{cases}
$$
And that for $(H_0,H_1'')$ is
$$
\phi_2(\mathbf X)=\begin{cases} 1 &,\text{ if }\frac{\sqrt n(\overline X-\theta_0)}{\sigma_0}<-z_{\alpha}
\\ 0 &,\text{ otherwise }
\end{cases}
$$
So the test functions $\phi^*$ and $\phi_1$ should coincide on the sets where $\phi_1$ is zero or one. Same goes for $\phi^*$ and $\phi_2$. Now suppose we observed a data $\mathbf X$ such that the observed value of $\frac{\sqrt n(\overline X-\theta_0)} {\sigma_0}$ exceeds $z_{\alpha}$. Then for such $\mathbf X$, we must have $\phi_1(\mathbf X)=1$ and $\phi_2(\mathbf X)=0$. This means that on the part of the sample space where $\frac{\sqrt n(\overline X-\theta_0)} {\sigma_0}>z_{\alpha}$, the test $\phi^*$ fails to coincide with both $\phi_1$ and $\phi_2$. Hence the contradiction.
This is pretty much the idea behind the nonexistence of a UMP test for $(H_0,H_1)$. Hence the LRT is not a UMP test; however it is a UMP unbiased (UMPU) test.
Best Answer
As in this question, the idea is directly comparing $f(X, Y) = XY - X(\tau)Y(\tau)$ and $g(X, Y) = |XY|(I(|X| \geq \tau) + I(|Y| \geq \tau))$ on different regions of $\Omega$.
On the region $[|X| < \tau, |Y| < \tau]$, $f(X, Y) = 0 \leq g(X, Y) = 0$.
On the region $[|X| < \tau, |Y| \geq \tau]$, $f(X, Y) = X(Y - Y(\tau)) = X(Y - \tau\operatorname{sign}(Y))$, $g(X, Y) = |XY|$. To see $f(X, Y) \leq g(X, Y)$ on this region, note that if $|Y| \geq \tau$, then $|Y - \tau\operatorname{sign}(Y)| \leq |Y|$.
On the region $[|X| \geq \tau, |Y| \geq \tau]$, $f(X, Y) = XY - \tau\operatorname{sign}(X)\tau\operatorname{sign}(Y)$, $g(X, Y) = 2|XY|$. It follows by $|\tau\operatorname{sign}(X)| = \tau \leq |X|$ and $|\tau\operatorname{sign}(Y)| = \tau \leq |Y|$ on this region that \begin{align} f(X, Y) \leq |f(X, Y)| \leq |XY| + |\tau\operatorname{sign}(X)| |\tau\operatorname{sign}(Y)| \leq |XY| + |XY| = g(X, Y). \end{align}
Now you should be able to finish the comparison on the remaining $1$ region.