Suppose $B$ is a standard Brownian motion and define $$f(x) \equiv \begin{cases}x & x \leq 3 \\ x +1 &x>3 \end{cases} $$
I want to show that $X_t \equiv f(B_t)$ is NOT right continuous with left limits, but I'm not 100% sure if my proof here is valid.
We proceed by contradiction. Suppose $X_t$ has right continuous paths and define $T = \inf\{t \ge 0: B_t > 3 \}$. It is commonly known that $T$ is finite almost surely, so that $X$ is right continuous at $T$ almost surely. We also know by the strong Markov property for Brownian motion that $B_{T+t} - 3$ is also a Brownian motion, and by Blumenthal's 0-1 law, $$\inf \{t \ge 0: B_{T+t} > 3\} = 0 = \inf\{t \ge 0: B_{T+t} < 3\} \quad \text{almost surely}$$
We also know that for Brownian motion, there exists a $\delta^* > 0$ such that $s, t \in [T, T+ \delta^*) \implies |B_t -B_s| < 1/2$.
Now fix $\delta > 0$ and note, by definition of the infimum, there exist $s, t \in [T, T+\delta \wedge \delta^*)$ such that $B_{T+t} >3, B_{T+s} <3$ so that $$|X_{T+t} - X_{T+s}| = |1 + B_t - B_s| \ge 1 - |B_t-B_s| \ge 1/2$$ and this contradicts the definition of right continuity at $T$, for there would have to exist a $\delta >0$ such that for every $t \in [T, T+\delta), |X_T - X_{T+t}| < 1/4$ so that $s, t \in [T, T+ \delta)$ implies $|X_{T+t} - X_{T+s}| < 1/2$.
I didn't consider anything to do with left limits so I just want to make sure this result is complete. Thanks for any feedback.