I'm having great difficulty trying to understand brownian motion. They require almost every path to be continuous, yet in proofs they assume all path are continuous. In quite a few questions I'm trying to prove, I need all path to be continuous too. Example: let $K\subseteq \mathbb{R}$ by closed, prove $T(\omega) = \inf\{t:B_t(\omega)\in K\}$ is a stopping time. My method is first let $U_\varepsilon = \bigcup_{x\in K} \mathcal{B}(x,\varepsilon)$, \begin{equation} \{T>t\}= \bigcap_{x\in[0,t]} B^{-1}_x(K^c) = \bigcup_n \bigcap_{q\in [0,t]} B^{-1}_q(U_{1/n}^c) \end{equation} where $q$ is rational. I need every $\omega$ to be continuous for the second equality to hold.
Discontinuous paths also defeat the purpose of stopping time, since they can leap across the enclosed boundary used to define $T$. This is not possible for discrete random walk $S_n$, no walk can leap across the boundary. What to do about the discontinuous path?
Even though the trajectories of a Brownian motion $B$ are assumed to be a.s. continuous, i.e. there exists a measurable set $A$ of probability one such that on $A$ every trajectory is continuous, you can always define another Brownian motion $\tilde B$ to be equal to $B$ on $A$ and identically $0$ otherwise. This way, $\tilde B$ has everywhere continuous trajectories and $B=\tilde B$ a.s.
Now, $\tilde T$ (hitting time of $K$ for $\tilde B$) is a stopping time, and that is what you have proven. Since $T=\tilde T$ a.s., if the filtration is complete, then it readily follows that $T$ is also a stopping time.
Completeness is usually a key condition for hitting times to be stopping times (see the Debut theorem, e.g. here).