Why is the probability that a continuous random variable takes any one specific value equal to 0? What would the intuitive explanation be?
Would it be because there are infinately many values and so the probability of any specific value is infinitely small hence we say 'close enough' to 0?
 A: A continuous random variable has the following property
$$ P( a \leq x \leq b) = \int_{a}^{b} f(x) \textrm{d}x $$
where the pdf of the RV is given by $f(x)$. 
In calculus we know that if the upper and lower limits of the integral are the same then it is $0$.
$$  P( c \leq x \leq c) = \int_{c}^{c} f(x) \textrm{d}x = 0 $$
You could probably justify this a few ways, but from the fundamental theorem of calculus we have that
$$  F(b) - F(a) = \int_{a}^{b} f(x) \textrm{d}x $$
so then the integral at a single point is
$$ F(c) - F(c) = 0 $$
A: In continuous case probabilities are defined as integrals. The probability of a single point is zero because you have an integral whose upper and lower limits are the same. 
The intuitive explanation is that you have uncountable infinitely many options with and you randomly pick one of them. What is the probability of choosing a specific one? 
The randomness among uncountable infinite number of choices makes it very unlikely to come up with the one that you like to choose. 
If you randomly pick a person what is the probability that he or she has exactly the same height as you have? 
A: Intuition
(Building on the MPW comment): If $P[X=x^*]=p>0$ at a certain point $x^*$ then 
\begin{align}
&\{X\leq x^*\} = \{X=x^*\} \cup \{X<x^*\}\\
&\\
\implies \quad &P[X\leq x^*] = p+P[X<x^*]  \quad \quad (Eq. 1)
\end{align}
You can see there is a positive gap of size $p$ between $P[X\leq x^*]$ and $P[X<x^*]$, which intuitively shows why the CDF function $P[X\leq x]$ jumps discontinuously at $x^*$.  
Thus, if the CDF has no discontinuous jumps then there can be no such point masses and so $P[X=x]=0$ for all $x \in \mathbb{R}$.

A full proof
Assume $P[X=x^*]=p>0$. Fix $x<x^*$ and notice that 
$$ \{X\leq x\} \subseteq \{X < x^*\} \implies P[X\leq x] \leq P[X< x^*]$$  Substituting this inequality into (Eq. 1) gives
$$ P[X\leq x^*] \geq p+ P[X\leq x] \quad \forall x < x^*$$
This holds for all $x<x^*$ and so it holds even when taking a limit as $x$ approaches $x^*$ from the left: 
$$ P[X\leq x^*] \geq p+\lim_{x\nearrow x^*}P[X\leq x] $$
So there is a positive gap of size (at least) $p$ between the value $P[X\leq x^*]$ and the limit of the CDF function $P[X\leq x]$ as $x$ approaches $x^*$ from the left. 
*In fact it can be shown that $\lim_{x\nearrow x^*}P[X\leq x]=P[X<x^*]$ and so from (Eq. 1) we see the gap is exactly $p$. The size of the discontinuous jump in the CDF function at $x^*$ is exactly the probability that $X=x^*$.
