You must show the set of Holder continuous functions is a countable union of nowhere dense sets in the metric space $C([0,1]).$ Here's a start: Show that $f(x)=x$ can be uniformly approximated by functions that are not Lipschitz. Hint: $\sqrt x$ is very close to $x$ on $[0,b]$ if $b$ is small enough.
Added later: I'll write $C$ for $C([0,1])$ and $\|\,\|$ for the sup norm. Let $H^a$ denote the Holder continuous functions on $[0,1]$ of order $a\in (0,1].$ Note that if $f \in H^{a_0},$ then $f\in H^a$ for $0<a<a_0.$
Theorem: Assume $f\in C,a>0$ and $\epsilon>0.$ Then there exists $g\in C\setminus H^a$ such that $\|f-g\|<\epsilon.$
Proof: Let's first treat the case $f(x)=x.$ For $b>0, b$ small, $0\le x^{a/2}-x \le b^{a/2}, x\in [0,b].$ Let's choose $b$ so that $b^{a/2}<\epsilon.$ Now define $g$ as follows:
$$ g(x) = \begin{cases} x^{a/2}, 0 \le x \le b \\
b^{a/2}, b \le x \le b^{a/2}\\
x, b^{a/2}\le x \le 1.\end{cases}$$
It's good to draw a picture. So informally, $g$ moves up above $x$ on $[0,b],$ but not by much. Then $g$ is constant on the next interval until it hits $f(x)=x.$ Then $g=f$ on the last interval. This $g$ is continuous, and the way we've set things up gives $\|f-g\|<\epsilon.$ And because $g(x)=x^{a/2}$ near $0,$ we have $g\not \in H^a.$
For the general $f\in C,$ let's recall that piecewise linear continuous functions are dense in $C.$ So it's enough to prove the theorem for piecewise linear $f.$ But now we're essentially in the special case above. After all, $f$ starts off as linear on some interval $[0,\delta].$ On this interval, we have $f(x) = f(0)+mx$ for some $m.$ Designing $g$ here is so much like the special case that I'll leave this final step to the reader. (End of proof)
Let's now define $E_{m,n} = \{f\in C : |f(y)-f(x)|\le n|y-x|^{1/m},x,y \in [0,1]\}.$ The set of all Holder continuous functions is then $\cup_{m,n \in \mathbb {N}} E_{m,n}.$ Furthermore, each
$E_{m,n}$ is closed in $C;$ that's a straightforward argument. The theorem shows that every $E_{m,n}$ has empty interior: $E_{m,n}$ couldn't contain a neighborhood of any $f\in C,$ because we know we can approximate $f$ as close as we want with functions $g \in C \setminus E_{m,n}.$ So the set of Holder functions is the countable union of closed nowhere dense sets, hence is of the first category in $C.$
Best Answer
Yes, when $x-y$ is small, the definition dictates that $f(x)-f(y)$ should be small in a certain way. This is what the definition of uniform continuity does as well, with its $\epsilon$ and $\delta$.
Maybe you should start with something simple and concrete: show that the function $f(x)=x^2$ on the interval $(0,1)$ satisfies $|f(x)-f(y)|\le 2|x-y|$.
Then proceed to something harder: show that $f(x)=\sqrt{x}$ is Hölder continuous with $\alpha=1/2$.
As for your original question, the best way to handle it is with Hölder's inequality. Road map: