By definition, $U_3 \subset \mathbb{R}P^2$ is open if and only if its preimage in $S^2$ is open (its preimage under the map $p : S^2 \to \mathbb{R}P^2$).

Hence, we want to prove $V_3$ is open in $S^2$. Because $V_3 = S^2 - \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_3 \ne 0\}$, and $S^2$ inherits the topology of $\mathbb{R}^3$, so if we can show an open set $W \subset \mathbb{R}^3$ such that $W \cap S^2 = V_3$, then $V_3$ is an open set. Then we can see if we define $W := \mathbb{R}^3 - \{(x_1,x_2,x_3) \in \mathbb{R}^3 \mid x_3 \ne 0\}$, it's open and $W \cap S^2 = V_3$, so $V_3$ is open in $S^2$ and $U_3$ is open in $\mathbb{R}P^2$.