Understanding $P(X=n)=0$ when $X$ follows a continuous distribution I have something that I can't get about continuous distribution. Let's say a variable $X$ follows a continuous distribution on $]0;1[$. Then when I pick a completely random number on $]0;1[$, say I have a number $n$. But, $P(X=n)=0$, so getting $n$ through this picking should be impossible. What's wrong in my thoughts?
 A: With continuous probabilities, it is necessary to suspend your intuition a little. A lot of our probability intuitions are related to finite sets. It takes some time and experience to understand how to extend those intuitions to the continuous case.
If $P(X=n)>0$ then the function $f(n)=P(X<n)$ can't have a continuous probability, because the difference between $f(n+\epsilon)$ and $f(n-\epsilon)$ is always at least $P(X=n)$.
It is better to consider continuous probabilities to measure the likelihood not of particular values, but the likelihood of a value being in a range. 
On some level, there is no such thing as actually picking a continuous random number from $0$ to $1$. It might seem like we can, but we can't. A continuous random variable is really a limit of a lot of finite random variables. It has a lot of the properties of finite random variables, but it fails some of our intuition in odd little ways.
As another answerer pointed out, in continuous random variables, probability zero doesn't mean "impossible." Consider it more like area - what is the area of a line segment? It is zero. But the line segment is not itself an empty set, it is just negligible when talking about area. If we pick a random point in a square, what is the probability that point will be on the diagonal? It will be zero, and that is precisely because the "area" of the diagonal line segment is zero.
A: 
so getting $n$ through this picking should be impossible

No, it is not "impossible". It is still "possible" even if it has zero probability. "Having zero probability" and "being impossible" are not the same things in the perspective of Probability Theory.
Let's consider your example. Let $X$ be uniformly distributed in the interval $[0,1]$. Then its PDF will be
$$ f_X(x) = \begin{cases} 1; & \text{if $0 \le x\le1$} \\ 0; & \text{otherwise.} \end{cases} $$
For example, we try to find the probability $P(X=0.6)$; that is
$$ P(X=0.6) = \int\limits_{x=0.6}^{x=0.6} f_X(x) dx = 0. $$
$P(X=0.6)$ is zero because the region of integration is zero lenght.

But there is a special case. The probability $P(X=k)$ will not be zero if there is an impulse at $x=k$ on the PDF of $X$.
For instance, consider a fair die. The PDF of outcomes are
$$ f_X(x) = \frac{1}{6}\delta(x-1) + \frac{1}{6}\delta(x-2) + \frac{1}{6}\delta(x-3) + \frac{1}{6}\delta(x-4) + \frac{1}{6}\delta(x-5) + \frac{1}{6}\delta(x-6) $$
where $\delta(x)$ is the unit impulse function on the origin.
For example, probability of seeing $5$ on the die will not be zero; it will be
$$ P(X=5) = \int\limits_{x=5}^{x=5} f_X(x) dx = \int\limits_{x=5}^{x=5} \frac{1}{6}\delta(x-5) dx = \frac{1}{6}. $$
