Why does the number of possible probability distributions have the cardinality of the continuum?

A probability on $\mathbb{R}$, be it continuous or not, is given by its CDF $x \mapsto\mathbb{P}(X \leq x)$. A CDF is right-continuous, and the set of right-continuous functions has the cardinality of $\mathbb{R}$. To see this, you can for instance argue that the values of such a function are given by its values at the rational points, so it has at most the cardinality of a countable product of copies of $\mathbb{R}$, which has the cardinality of $\mathbb{R}$ as well.

To expand on the AC/CH question, Raoul's argument does not depend on either of these, since you can give an explicit injection from real-valued sequences $x_1,x_2,\ldots$ to $\mathbb R$ (and there is an explicit bijection between $\mathbb Q$ and $\mathbb N$, so between $\mathbb R^{\mathbb Q}$ and $\mathbb R^{\mathbb N}$). To do this, write each value as an decimal (converting $0.1999...$ to $0.2$, etc.). Then form a new infinite decimal as follows: digits in odd places are the digits of $x_1$, in order; those in places $\equiv 2$ mod $4$ are digits of $x_2$; those in places $\equiv 4$ mod $8$ digits of $x_3$; and so on. Since infinitely many digits of $x_1$ are not $9$, the same is true of the decimal we obtain by this process, and you can easily recover the digits of each $x_i$ from the final decimal, so this is an injection.