Can an observed event in fact be of zero probability? Can an observed event in fact be of zero probability?
Of course, I know that there exist non-empty events of zero probability. Mu question is the reverse: given that we have observed an event (and we have no other information about it, just the fact that it has been observed), is it possible that the event has in fact zero probability? Or does an observation necessarily mean that the probability is strictly positive?

Example context for the question:
Suppose $x_i$ (countably many) are i.i.d on $[0,1]$, but we do not know the distribution they come from. It may be uniform on $[0,1]$, may be discrete, may be any legitimate distribution (discrete or continuous). Just imagine we have some sort of a machine that shows us one by one a list of randomly drawn numbers between $0$ and $1$. We are comparing the observed numbers one by one to some special number chosen beforehand, for example $\frac12$. Now, given that at some iteration we have observed that special number at least once, does that mean that $\frac12$ has some positive probability under that (unknown) distribution? And if the probability can be zero, can we nevertheless say that we will necessarily observe $\frac12$ again later, if we continue the experiment ad infimum?
Also, disregard the "real world limitations" such as an inability to produce truly uniformly distibuted numbers, or rounding errors or any such thing.
 A: Your question is not a mathematical question, since it depends on the interpretation of probability. It also depends on precisely what you mean by "given that we have observed...".
For example, subjective Bayesians think of probability theory as theory of how (idealized) rational agents should update their beliefs in light of new observations. It is part of this idealized framework that when an event is observed, the probability measure gets updated in such a way that the observed event gets probability 1. IOW, the a posteriori probability of an observed event is not only non-zero, but equal to 1. On the other hand, the a priori probability (the probability before the update) could be anything. Normally, Bayesians adopt additional rationality principles such as always assigning non-zero a priori probabilities to non-empty events in finite sample spaces. But modelling idealizations or misjudgements about the appropriate sample space might still amount to assigning a zero probability to events that are possible.
In an experiment where a sugar cube with faces labeled 1,2,3,4,5,6 is thrown exactly once onto a wet surface, Bob might choose the sample space {1,2,3,4,5,6} and Alice might choose the sample space {1,2,3,4,5,6,edge}. If the sugar cube ends up balancing on an edge, Alice could update her probabilities using Bayes theorem (if she assigned edge some small but non-zero probability). Bob, on the other hand, would've observed an event that he had effectively assigned a vanishing a priori probability. So he would need to update using some other rule than Bayes rule, since he also needs to expand his sample space.
There are many other interpretations of probability than subjective Bayesianism, and the detailed answer to your question may vary depending on which interpretation you prefer. This is why it is not a mathematical question.
EDIT: Regarding the added clarification to the original question, it is certainly possible (and typical!) to have a zero probability of $x_i = 1/2$ in the continuous case. In the discrete case, where the probability mass is concentrated on some finite or countably infinite discrete subset of [0,1], I am not sure. If, say, one has observed that $x_1 = \frac{1}{2}$, then intuitively it does seem plainly irrational to base predictions about $x_k$, $k > 1$, on a probability measure that assigns $P(x_k=\frac{1}{2}) = 0$. But I don't have any formal argument to support this...
A: 
is it possible that the event has in fact zero probability?

Yes. For a situation where this always happens, assume that one observes a random number $x$ drawn from the uniform distribution on $(0,1)$. Then the probability to observe $x$ is zero. 
(Proof: For every interval $I\subseteq(0,1)$ the probability to observe a number in $I$ is the length of $I$. For every positive $\varepsilon$, there are intervals  $I\subseteq(0,1)$ which contain $x$ and have length less than $\varepsilon$ hence the probability to observe exactly $x$ is less than $\varepsilon$, for every positive $\varepsilon$, QED.)
Edit: In a discrete space $\{x_i\mid i\in\mathbb N\}$, the probability to observe $x_i$ is, by hypothesis, some positive $p_i$ hence the above applies only to continuous distributions (or at least, partially continuous).
A: If I understand your question correctly, this is impossible due to Bayes rule:
$$
P(V|S) = \frac{P(S|V)P(V)}{P(S)}
$$
where $V$ is the event. Clearly LHS is greater than zero (let's say $\epsilon$) because you have observed it. If its probability is 0 though, then $P(V)=0$ and RHS is 0, and $\epsilon \neq 0$.
