This question really pertains to any discrete time continuous-valued, stationary stochastic process on the real line, but the Gaussian process will be adequate for this question.
I have this confusion about zero-probability events. If we observe a discrete-time Gaussian process for any finite length of time, the probability that it will hit a particular real number is $0$. However, this does not imply that it is impossible for the process to take on this value, as impossible events are only a subset of zero-probability events.
Now, instead of watching the process for a finite period of time, lets say that you are going to make a bet that at some time in the future, the Gaussian process will take a particular (real-number) value at least once.
So here's my confusion: it seems that I can argue this both ways. I can say that since it is a zero probability event, there are an infinite number of other numbers it can visit, so its zero. On the other hand, with infinite time, the empirical CDF that results from the Gaussian process sample path will converge to the actual CDF of a Gaussian RV, so it converges to a distribution that has visited all points in the real line.
Can someone provide some insight into which of the above is more accurate when $T=\infty$?
As a related question: It seems that the continuous time, nonstationary version of this (Brownian Motion) is actually easier to think about, as the sample paths are continuous and hence must cover all values between its most extreme endpoints, which is asymptotically the entire real line.
Let $\xi_n$ denote the value of the process at time $n$, then the random set $\{\xi_n:n \in\mathbb{N}\}$ will always be countable. It can never cover the entire real line, which is uncountable. So the answer to the question in your title is negative.
As for the bet that a particular number $x\in\mathbb{R}$ will be visited: Let $A_n=\{\xi_n\neq x\}$, so that $P(A_n)=1$ for all $n$. Because the events $A_1,A_2,\ldots$ are independent, $P(\cap_{n\in\mathbb{N}} A_n)=\prod_{n\in\mathbb{N}} P(A_n)$, hence $x$ is almost surely never visited.
The situation for Brownian motion is as you described it.