I'll elaborate on my comment. I claim that the statement "The probability that a continuous random variable takes on a specific value actually equal zero?" is false. I'll stick with the definition that a continuous random variable takes values in an uncountable set, or, to be more precise, that no countable subset has full measure. It is the one used by Davitenio, and in the intro of this Wikipedia article.
Take your favorite real-valued continuous random variable; call it $X$. Flip a well-balanced coin. Define a random variable $Y$ by:
- If the coin shows heads, then $Y=X$;
- If the coin shows tails, then $Y=0$.
The random variable $Y$ has the same range as $X$: any value taken by $X$ can be achieved by $Y$ provided that the coin shows heads. Hence, it is continuous. However, with probability at least $1/2$, we have $Y=0$, so that one specific value has non-zero probability.
The good notion here would be the notion of non-atomic measure. An atom is a point with positive measure, so a random variable which doesn't take any specific value with positive probability is exactly a random variable whose image measure is non-atomic. This is a tautology.
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Another definition of "continuous random variable" is a real-valued (or finite-dimensional-vector-space-valued) random variable whose image measure has a density with respect to the Lebesgue measure. Yes, even Wikipedia gives different definitions to the same object.
If $X$ is a continuous random variable with this definition, then is a function $f$, non-negative and with integral equal to $1$, such that for any Borel set $I$, we have $\mathbb{P} (X \in I) = \int_I f(x) dx$. Since a singleton has zero Lebesgue measure, we get $\mathbb{P} (X = x) = 0$ for all $x$.
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My take on the subject (warning: rant): I really, really don't like the use of "continuous random variable", and more generally the use of "continuous" in opposition to "discrete". These are the kind of terms which are over-defined, so that you can't always decide what definition the user has in mind. Even if it is quite bothersome, I prefer the use of "measure absolutely continuous with respect to the Lebesgue measure", or with some abuse, "absolutely continuous measure", or "measure with a density". With even more abuse, "absolutely continuous random variable". It is not pretty nor rigorous, but at least you know what you are talking about.
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PS: As for why your proof does not work, Joriki's answer is perfect. I would just add that the formula
$$\mathbb{P} (X = x) = \frac{\# \{ \text{favorable outcomes} \} }{\# \{\text{possible outcomes}\}}$$
only work with finite probability spaces, and when all the outcomes have the same probability. This is what happens when you have well-balanced coins, non-loaded dices, well-mixed card decks, etc. Then, you can reduce a probability problem to a combinatorial problem. This does not hold with full generality.