How to explain why the probability of a continuous random variable at a specific value is 0? Consider X as a continuous random variable which can assume any value in [0, 1]. It is known that P(X=x)=0 where P is the probability density function. I want to understand this intuitively.
The math insight article helps me somewhat:

In other words, the probability that the random number X is any
  particular number x∈[0,1] (confused?) should be some constant value;
  let's use c to denote this probability of any single number. But, now
  we run into trouble due to the fact that there are an infinite number
  of possibilities. If each possibility has the same probability c and
  the probabilities must add up to 1 and there are an infinite number of
  possibilities, what could the individual probability c possibly be? If
  c were any finite number greater than zero, once we add up an infinite
  number of the c's, we must get to infinity, which is definitely larger
  than the required sum of 1. In order to prevent the sum from blowing
  up to infinity, we must have c be infinitesimally small, i.e., we must
  insist that c=0.

But, what if I choose, c=1/N (where N is that large number) for the uniform case? The sum will still be 1 as far as I can understand. For the non-uniform case, I can pick some 0's and others non-zeros and still be theoretically able to get a sum of 1 for all the possible values.
Anything that helps me understand this clearly will be of immense help. The answer here helps but I still don't get it.
 A: A continuous random variable can realise an infinite count of real number values within its support -- as there are an infinitude of points in a line segment.
So we have an infinitude of values whose sum of probabilities must equal one.  Thus these probabilities must each be infinitesimal.  That is the next best thing to actually being zero.  We say they are almost surely equal to zero.
$$\Pr(X=x) = 0 \text{ a.s.}$$
( To have a sensible measure of the magnitude of these infinitesimal quantities, we use the concept of probability density, which yields a probability mass when integrated over an interval. This is, of course, analogous to the concepts of mass and density of materials. )
$$f_X(x) = \frac{\mathrm d}{\mathrm d x}\Pr(X\leq x)$$


For the non-uniform case, I can pick some 0's and others non-zeros and still be theoretically able to get a sum of 1 for all the possible values.

You are describing a random variable whose probability distribution is a mix of discrete (massive) points and continuous intervals.  This has step discontinuities in the cumulative distribution function.
$\Pr(X\leq x) = \begin{cases} 0 & x < 0 \\ 0.25 & x=0 \\ 0.25+x/4 & 0< x< 1/2 \\ 0.75 & x=1/2 \\ 0.5+x/4 & 1/2< x< 1 \\ 1 & x\geq 1\end{cases}\\[2ex] \Pr(X=x) = \begin{cases} 0 & x<0 \cup x> 1 \\ 0.25 & x=0 \cup x=1/2 \cup x=1 \\ 0\text{ a.s.} & 0<x<1/2 \cup 1/2<x< 1 \end{cases}$
A: The situation is easier to formalize when we say that there is no uniform distribution on a countably infinite set $S$. This is because if there were, then $P(X=x)=c>0$ for every $x$, and now
$$P(X \in S)=\sum_{x \in S} c = +\infty.$$
This follows from the property of countable additivity of probability, which is usually treated as an axiom. I think that at least finite additivity of probability should be intuitively obvious, and countable additivity is a straightforward extension.
In the uncountable case, we need one more result, which can be proven from the countable additivity axiom: if $A \subset B$ then $P(A) \leq P(B)$; so now if $S$ is uncountable, and we are to assign a uniform distribution to it, then we can extract a countably infinite subset $C$. Then 
$$P(X \in S) \geq P(X \in C) = \sum_{x \in C} c = +\infty$$
as before.
A: Use the fact that if $A \subset B$, then $p(A) \le p(B)$.
If $X$ is a continuous random variable, this means that the cdf is continuous, that is $f(\alpha) = p(\{\omega | X(\omega) \le \alpha \})$ is continuous.
It follows that $p((a,b]) = f(b)-f(a)$.
Suppose $a < x \le b$. Then we have $p((a,b]) = f(b)-f(a)$.
Since $\{x\} \subset [a,b]$ for all such $a,b$, we have $p(\{x\}) \le f(b)-f(a)$ for all such $a,b$. Hence $p(\{x\})=0$, since $f$ is continuous.
