Let $X_t$ be a continuous martingale on $\mathbb{R}.$ From continuity we have that $X_t$ is uniformly continuous on closed interval let's say $[0,t]$. Uniformly continuous function on some set is bounded. Is't also true for $X_t$? I mean, \begin{equation} \forall \omega \sup_{0 \leq s \leq t} |X_t(\omega)| < M. \end{equation}
For example if I assume that $X_t$ is a standard Brownian motion then $X_t \sim N(0,t)$ and of course $X_t$ is unbounded. But from my reasoning above it's should be.
I would really appreciate any hints or tips.
Careful, your inequality should be:
$$\sup_{0 \leq s \leq t} |X_t (\omega )| < M_\omega$$
That is, by continuity, each sample path can be bounded on compact intervals, with the bound $M_\omega$ depending on each sample path. This does not imply that the entire process is bounded. As you mentioned, Brownian motion is a counterexample to this claim.