Here is an excerpt from "An Elementary Introduction to Mathematical Finance" by Sheldon Ross, 3rd edition:
I understand this is not meant to be rigorous, but I'm having trouble understanding conceptually what are the implications of $X(t)$ being a continuous function of $t$.
I know a random variable is a function from the sample space to the real line. Given $t \geq 0$ and an arbitrarily small $h$, we have that $X(t+h)$ is "close" to $X(t)$. Does this mean these random variables share the same sample space $\Omega$? In other words, does this mean for each outcome $\omega \in \Omega$, $X(t+h)(\omega)$ be "close" to $X(t)(\omega)$?
I know that Brownian motion can be used to model price movements, so I'm trying to imagine this in terms of a plot of price history being an unbroken curve. The way I'm thinking of this right now is that given an outcome $\omega$, the collection $X(t)(\omega)$ can be plotted for $t \geq 0$ and we will see a continuous curve. Please let me know if this interpretation is correct or if I'm totally off the mark.
Yes you are right, it means that $P(t \mapsto X(t) \text{ is continuous}) = 1$. In other words, for $\omega$ in a set of probability $1$, the map $t \mapsto X(w)(t)$ is continuous.
Meanwhile, the author proved that for each $t$, $X(t + h) \to X(t)$ in $L^2$ as $h \to 0$, i.e. $E((X(t + h) - X(t))^2) \to 0$ as $h \to 0$. There is a theorem (https://en.wikipedia.org/wiki/Kolmogorov_continuity_theorem) that shows that if you have moment bounds $E(|X(t + h) - X(t)|^{\alpha}) \leq K|h|^{1 + \beta}$, then there is a modification $\tilde{X}$ of the process $X$ that has continuous paths. Modification means that for each $t$, $P(\tilde{X}(t) = X(t)) = 1$. Since $X$ and $\tilde{X}$ have the same finite-dimensional distributions, we study $\tilde{X}$ instead of $X$. Though, it isn't necessarily true that $P(\tilde{X}(t) = X(t) \text{ for all } t) = 1$.