Zero probability and impossibility
Solution 1:
Two Schools
I think the crux of the matter is what probability actually is:
- The Bayesian view - probabilities are measures of (personal) confidence or belief, so it's quite obvious why an event with probability zero is not the same thing as an impossible event. But perhaps this isn't such a satisfactory answer.
- The frequentist view - probabilities are the asymptotic frequency of events as the number of independent trials tends to infinity. Here again wee see that something that happens with probability zero is not the same as something impossible; it's just something that happens so infrequently that the numerator in $\dfrac{\text{occurences}}{\text{trials}}$ is dominated by the denominator.
Technically Speaking
Putting aside such philosophical matters, there's also a technical matter to be discussed here. Under the usual measure-theoretic formulation of probability theory, we have a sample space $\Omega$ and a family $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ of events (measurable subsets of $\Omega$), and the probability of an event $A \in \mathcal{F}$ is its measure $\mathbb{P}(A)$. There is nothing in the axioms of measure theory which say that a non-empty set must have a non-zero measure; and if we interpret $\mathcal{F}$ as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability.
Example
To give a concrete example, consider a random variable $X$ which is uniformly distributed on the interval $[0, 1]$. Although $\mathbb{P}[X \in (a, b)] = b - a$ for all $(a, b) \subset [0, 1]$, the axioms of probability force us to conclude that $\mathbb{P}[X = x] = 0$ for any individual $x \in [0, 1]$: for if $\mathbb{P}[X = x] = \varepsilon > 0$, because $X$ is uniformly distributed, by additivity of the probabilities of disjoint events, we'd be forced to conclude that $[0, 1]$ contains at most $\frac{1}{\varepsilon}$ (a finite number!) points, which is absurd.
Solution 2:
Zero probability isn't impossibility. If you were to choose a random number from the real line, 1 has zero probability of being chosen, but still it's possible to choose 1.