Intuition behind why continuous random variables cannot take a particular value? It was iterated over and over again in our probability class that continuous random variables CANNOT have a probability mass function, instead they NEED TO have a density function. I am trying to find some intuition behind this. Any help would be greatly appreciated.
Additionally the following facts were mentioned:

*

*The probability distribution function of some continuous random variable may take a value larger than 1. Why/How?


*In the case of a continuous random variable, $P(X=x) \ne p_{x}(x)$. Why?
 A: If there is some $c\in \mathbb{R}$ such that $\Pr [X=c]>0$ then the distribution function of $X$ at $c$ is discontinuous. To see this we will show the contrapositive statement: if $F_X$ will be continuous at $c$ then we will show that $\Pr [X=c]=0$.
First, as by assumption $F_X$ is continuous at $c$, then
$$
\lim_{x\to c}F_X(x)=F_X(c)\tag1
$$
Also we have that
$$
\Pr [X=c]= \Pr [X\leqslant c]-\Pr [X<c]\tag2
$$
But as a probability is a measure then
$$
\begin{align*}
\Pr [X<c]&=\Pr [X^{-1}((-\infty ,c))]\\
&=\Pr \left[\bigcup_{n\in \mathbb N }X^{-1}((-\infty ,c-1/n])\right]\\
&=\lim_{n\to\infty}\Pr [X^{-1}((-\infty ,c-1/n])]\\
&=\lim_{n\to\infty}F_X(c-1/n)\tag3
\end{align*}
$$
Thus from (1) and (3) we find that if $F_X$ is continuous at $c$ then $\Pr [X<c]=\Pr [X\leqslant c]$, and consequently from (2) we have that $\Pr [X=c]=0$.∎
A: We know from the probability axioms that the total probability over a sample space is $1$. For a continuous random variable, this corresponds to the area to the left of a point under the curve. So, the probability $P(X=x)$ is the area of an infinitely small strip with some height, and so the area is $0$.
The way to formalise this is to consider the probability $P(X-\epsilon<x<X+\epsilon)$.
As for your first question, can you elaborate further with a specific example?
A: 
The probability distribution function of some continuous random variable may take a value larger than 1. Why/How?

For the same reason that you can drive somewhere one mile away while going at speeds faster than 1 mph.
The same general idea applies to the rest of the questions.
A: I'll elaborate on the idea of "almost never" in probability - this refers to an event that occurs with probability 0, but is not impossible. Suppose we want to pick a random number from the uniform distribution on [0, 1]. There are an infinite number of choices, so the odds that we will pick exactly 0.5 (or any other numer) is zero. The fact remains, however, that 0.5 is in the range we're selecting from, so it is actually possible to pick it. A continuous random value does take on a particular value, despite the fact that the likelihood of picking any particular value is actually zero. If you throw a dart at the number line in the [0, 1] range, you have zero likelihood of hitting any particular value with infinite precision, but the dart still must land somewhere.
A: Maybe this will help clarify things:
Suppose you tried to use a mass function for a continuous r.v.  This is a function that assigns nonzero probability to some basic, atomic outcomes.  There is an uncountably infinite number of outcomes that have values, though, given a continuous distribution.
How could you write a summation to define the expectation?  There are not enough numbers in the infinite series of integers that would index a summation.
OK, but suppose, magically (doesn't really make sense, but let's try it out for the sake of a reductio) you could write the expectation as a sum.  Since there is an infinite number of values, each with equal probability (otherwise the distribution will not be continuous), all values have zero probability, in which case the total probability is zero, or else the probabilities are nonzero, in which case the total probability is infinite.  But the sum must be equal to 1.
With a density function, you can "sum" using an integral, though. Each possible value of the r.v. has zero probability, but regions of values have nonzero probability.  That's why the expectation of a continuous r.v. can be written as an integral over a density function.
