For an example which is a.s. but not m.s. continuous, take your time interval to be $[0, \infty]$, and let $X_t$ be a standard one-dimensional Brownian motion started at 0 and stopped the first time it reaches 1. (That is, if $B_t$ is a standard Brownian motion, take $T = \inf\{t > 0 : B_t = 1\}$ and $X_t = B_{t \wedge T}$.) Since Brownian motion is recurrent, we have $X_t \to 1$ a.s. as $t \to \infty$, and so by setting $X_{\infty} = 1$ we get an a.s. continuous stochastic process on $[0,\infty]$.
However, $X_t$ is not m.s. continuous. If it were, then by Cauchy-Schwarz we would have $E[X_t] \to E[X_\infty] = 1$ as $t \to \infty$. But $X_t$ is a martingale and so $E[X_t] = 0$ for all $t \in [0,\infty)$.
If you don't like using $[0,\infty]$ as your time interval, then apply a time change: let $Y_t = \begin{cases} X_{t/(1-t)}, & t < 1 \\ 1, & t \ge 1.\end{cases}$ Now $Y_t$ is a.s. continuous but $E[Y_t] = 0$ for $t < 1$, $E[Y_t] = 1$ for $t \ge 1$. Note that $Y_t$ is a standard example of a local martingale which is not a martingale.