$(E,d)$ is a metric space. For every $r \in \mathbb{R}_+,$ let $X_r:(\Omega,\mathcal{Q}) \to (E,\mathcal{B}(E))$ be a family of random variables such that for every $w \in \Omega,$ the function $r \to X_r(w)$ is continuous on $\mathbb{R}_+.$ Let $G$ be a closed of $E$ and fix $u \in \mathbb{R}_+.$
Considering, for $w \in \Omega,\theta(w)=\inf\{r \in \mathbb{R}_+,X_r(w) \in G\},$ why are we allowed to write $$\{w \in \Omega,\theta(w) \leq u\}=\bigcup_{r \in [0,u]}\{w \in \Omega,X_r \in G\}?$$ In particular, why the inclusion $\subset$ holds true ?
Suppose $\omega$ belongs to the left side. There is a sequence $\{r_n\}$ in $\{r\in \mathbb R_{+}: X_r (\omega) \in G\}$ such that $r_n$ decreases to $\theta (\omega)$. We have $X_{r_n} (\omega) \in G$ for all $n$. By continuity of paths and closedness of $G$ we get $X_{\theta (\omega)} (\omega) \in G$. Take $r=\theta (\omega)$. Then $X_r (\omega) \in G$ and $r \leq u$. Hence $\omega$ belongs to the right side.