Is it possible to demonstrate that md5(x) != x for any x? I am looking for an easy to follow explanation, if possible, that demonstrates/proves the validity (or not!) of this assertion:

for any X, md5(X) != X   (being X any string of 32 hex characters)
It is not known whether there is a fixed point or not. Since MD5 gives you 128 bits, you would need an input of 128 bits as well. That is you are really considering a function from the set of 128-bit strings to 128-bit strings. Assume that such a function is chosen at random. Then we can calculate the probability that the function has a fixed point. The answer is stated in this answer: https://stackoverflow.com/questions/235785/is-there-an-md5-fixed-point-where-md5x-x. To unpack just a bit more:

Say you have any of the $2^{128}$ possible inputs. Then the probability that this is a fixed point is $(1/2)^{128}$ since, for example, the first bit would have to map to the first bit and so on. So the probability that the input is not a fixed point is $1 - (1/2)^{128}$. So the probability that no input will be a fixed point is 

$(1 - (1/2)^{128})^{(2^{128})}$

because there are $2^{128}$ possible input. 

The probability that there is a fixed point is therefore $1 - (1 - (1/2)^{128})^{(2^{128})}$. If you calculate this probability you get approximately 63.21 %.

So it is actually likely that there is a fixed point for MD5 or any other hash function for that matter. 

Note that this doesn't prove that there is a fixed point, but it just says something about the likelihood.

The only way to demonstrate/prove whether or not there is a fixed point would be to compute all hashes of all 128 bit strings and compare them to the output.  But, as notes in the comments above, this is not very practical.