Your argument has an error: $\sigma_B = t$ does not imply $X_t \in B$, only $X_t \in \overline{B}$.
Consider the following example. Let $\mathcal{X}$ be the real line and let $Z$ be a fair coin flip, so $Z = \pm 1$ with probability $1/2$. Set $X_t = tZ$. So this process flips a coin at time 0 to decide whether to move left or right, then moves in that direction at unit speed from then on. Let $\mathcal{F}_t = \sigma(X_s : s \le t)$ be the natural filtration. Note that $X_0 = 0$ surely so $\mathcal{F}_0$ is trivial. (This filtration is not right continuous.) Let $B = (0, \infty)$ be the open right half-line. Now if the coin $Z$ comes up $+1$, then $X_t$ is in $B$ for all positive times, hence $\sigma_B = 0$; but if $Z = -1$ then $X_t$ is never in $B$, so $\sigma_B = \infty$. Hence the event $\{\sigma_B \le 0\}$ equals the event $\{Z = 1\}$ which is a nontrivial event and hence not in $\mathcal{F}_0$. So $\sigma_B$ is not a stopping time.
(This also points out your error: even on the event $\{\sigma_B = 0\}$ we have $X_0 = 0 \notin B$.)
So we cannot drop the assumption that the filtration is right continuous.
You are correct that we do not need to also assume the filtration is complete, if we are assuming that the process is surely right-continuous. But in most cases, we only assume that the process is almost surely right continuous. In that case, all these computations need to throw in the null event that the process is not right continuous, and without completeness you cannot be sure that this event is in all the relevant $\sigma$-fields.