'Amount' of nowhere-differentiable functions in $C([0,1])$?

This comes down to different variants of the phrase "almost all".

Some of the more common uses of "almost all" are

• On a co-countable set.
• On a conull set (with respect to some measure), i.e. the measure theoretic version.
• On a dense set, i.e. the topological version.
• On a comeager set (with respect to some topology), i.e. the finer topological version (for nice enough spaces).

Some of these play well with others, some are stronger than others, and which one is more useful depends on the problem at hand.

For example, being null and being meager don't play nice with each other, as you can write $\mathbb{R}=N\cup M$, with $N$ null, $M$ meager and $N\cup M=\emptyset$. So, being small in one sense tells you absolutely nothing about being small in another set.

On a similar note to what you state, you can argue that every number is irrational because the complement is countable (while $\mathbb{R}$ is not). The same works to show that almost every number is transcendental. Incidentally, this is the easiest proof of the existence of transcendental numbers, and is again an example of "to construct an object with the property X, we shows that the set of objects without property X is 'small', and that the entire space is not 'small'". The same thing is done when trying to show that $L^1$ functions have Lebesgue points, you actually show that almost every point (with respect to Lebesgue measure) is a Lebesgue point.

In the case of non-differentiable functions, they form a comeager set, which is stronger than mere density (for nice enough spaces). For example, the intersection of two comeager sets is comeager, which is patently false for dense sets.

To reiterate, the notion of "size" which you want to use with will depend on what you're trying to do. If you're working in set theory, for example, club sets fill the purpose of largeness.