Consider the flow on $\Omega={\mathbb R}^2$ given by $$\phi_t(x,y)=(e^tx, e^{-t}y).$$ Take $U_1$ equal to the open upper half-plane ($y>0$) and $U_2$ equal to the open right half-plane ($x>0$). I leave it to you to verify that the flow is proper on both $U_1, U_2$ and is not proper on their union.

This is @Ruy, editing @Moishe's post in order to give a few more details to their nice counter-example.

1. The flow is proper on the open right half-plane because it is conjugate to the very simple flow $$ \psi _t(x, y) = (x+t, y), $$ on ${\mathbb R}^2$, via the diffeomorphism $$ (x, y)\in (0, +\infty )\times {\mathbb R}\ \mapsto \ \big (\ln(x), xy\big )\in {\mathbb R}^2. $$ A similar reason applies to prove the properness of the flow on the open upper half-plane.

2. The flow is not proper on the union of the two half-planes mentioned above because the sequence $\{v_n\}_{n\in {\mathbb N}}$, given by $$ v_n=(e^{-n}, 1), $$ converges to $(0,1)$, and setting $t_n=n$, then $$ \varphi _{t_n}(v_n) = (1, e^{-n})\to (1,0), $$ and yet the $t_n$ are unbounded.

Consider the flow on $\Omega={\mathbb R}^2$ given by $$\varphi_t(x,y)=(e^tx, e^{-t}y).$$ Take $U_1$ equal to the open upper half-plane ($y>0$) and $U_2$ equal to the open right half-plane ($x>0$). I leave it to you to verify that the flow is proper on both $U_1, U_2$ and is not proper on their union.

Edit.

1. The flow is proper on the open right half-plane because it is conjugate to the linear flow $$ \psi _t(x, y) = (x+t, y), $$ on ${\mathbb R}^2$, via the diffeomorphism $$ (x, y)\in (0, +\infty )\times {\mathbb R}\ \mapsto \ \big (\ln(x), xy\big )\in {\mathbb R}^2. $$ A similar reasoning implies properness of the flow on the open upper half-plane.

2. The flow is not proper on the union $U=U_1\cup U_2$ because the sequence $\{v_n\}_{n\in {\mathbb N}}\in U$, given by $$ v_n=(e^{-n}, 1), $$ converges to $(0,1)$, while for $t_n=n$, we get $$ \varphi_{t_n}(v_n) = (1, e^{-n}) $$ which converges to $(1,0)\in U$ as $n\to\infty$.

Consider the flow on $\Omega={\mathbb R}^2$ given by $$\phi_t(x,y)=(e^tx, e^{-t}y).$$ Take $U_1$ equal to the open upper half-plane ($y>0$) and $U_2$ equal to the open right half-plane ($x>0$). I leave it to you to verify that the flow is proper on both $U_1, U_2$ and is not proper on their union.

Consider the flow on $\Omega={\mathbb R}^2$ given by $$\phi_t(x,y)=(e^tx, e^{-t}y).$$ Take $U_1$ equal to the open upper half-plane ($y>0$) and $U_2$ equal to the open right half-plane ($x>0$). I leave it to you to verify that the flow is proper on both $U_1, U_2$ and is not proper on their union.

This is @Ruy, editing @Moishe's post in order to give a few more details to their nice counter-example.

1. The flow is proper on the open right half-plane because it is conjugate to the very simple flow $$ \psi _t(x, y) = (x+t, y), $$ on ${\mathbb R}^2$, via the diffeomorphism $$ (x, y)\in (0, +\infty )\times {\mathbb R}\ \mapsto \ \big (\ln(x), xy\big )\in {\mathbb R}^2. $$ A similar reason applies to prove the properness of the flow on the open upper half-plane.

2. The flow is not proper on the union of the two half-planes mentioned above because the sequence $\{v_n\}_{n\in {\mathbb N}}$, given by $$ v_n=(e^{-n}, 1), $$ converges to $(0,1)$, and setting $t_n=n$, then $$ \varphi _{t_n}(v_n) = (1, e^{-n})\to (1,0), $$ and yet the $t_n$ are unbounded.