This is one of the well know examples of Weierstrass function. Hardy studied Hölder-continuity of such functions in Weierstrass's nondifferentiable function, G.H. Hardy, Trans. Amer. Math. Soc., 17 (1916), 301–325.. Here is a proof of Hölder-continuity for your case.
Theorem. Let $0<a<1$, $b>1$ and $ab>1$ then the function
$$
f(x)=\sum\limits_{n=1}^\infty a^n\cos(b^n x)
$$
is $(-\log_b a)$-Hölder continuous.
Proof.
Consider $x\in\mathbb{R}$ and $h\in(-1,1)$, then
$$
f(x+h)-f(x)=
\sum\limits_{n=1}^\infty a^{n}(\cos(b^n(x+h))-\cos(b^nx))=
$$
$$
-2\sum\limits_{n=1}^\infty a^{n}\sin(2^{-1}b^n(2x+h))\sin(2^{-1}b^{n}h)=
$$
Since $b>1$ and $h\in(-1,1)$ there exist $p\in\mathbb{N}$ such that $2^{-1}b^{p}|h|\leq 1<b^p|h|$, so
$$
|f(x+h)-f(x)|\leq
2\sum\limits_{n=1}^\infty a^{n}|\sin(2^{-1}b^{n}(2x+h))||\sin(2^{-1}b^{n}h)|\leq
2\sum\limits_{n=1}^\infty a^{n}|\sin(2^{-1}b^{n}h)|=
$$
$$
2\sum\limits_{n=1}^p a^{n}|\sin(2^{-1}b^{n}h)|+
2\sum\limits_{n=p+1}^\infty a^{n}|\sin(2^{-1}b^{n}h)|\leq
2\sum\limits_{n=1}^p a^{n}|2^{-1}b^{n}h|+
2\sum\limits_{n=p+1}^\infty a^{n}=
$$
$$
\frac{ab|h|}{ab-1}(a^p b^p-1)+\frac{2a}{1-a}a^p\leq
\frac{ab|h|}{ab-1}a^p b^p+\frac{2a}{1-a}a^p
$$
Since $2^{-1}b^{p}|h|\leq 1<b^p|h|$ and $0<a<1$, then we have $b^p|h|<2$ and $a^p\leq |h|^{-\log_b a}$. Hence
$$
|f(x+h)-f(x)|\leq
\frac{ab|h|}{ab-1}a^p b^p+\frac{2a}{1-a}a^p\leq
\frac{2ab}{ab-1}a^p +\frac{2a}{1-a}a^p\leq
$$
$$
\left(\frac{2ab}{ab-1} +\frac{2a}{1-a}\right)|h|^{-\log_b a}=
\frac{b-1}{(ab-1)(1-a)}|h|^{-\log_b a}
$$
This means that $f$ is $(-\log_b a)$-Hölder continuous.
Corollary. For $\alpha\in(0,1)$ the function
$$
f(x)=\sum\limits_{n=1}^\infty 2^{-n\alpha}\cos(2^n x)
$$
is $\alpha$-Hölder continuous.
Proof Apply previous theorem with $a=2^{-\alpha}$ and $b=2$.
As for the proof of nowhere differentiability, I don't know a short proof. The problem is that the standard Weierstrass argument is not applicable here - parameters $a$, $b$ must satisfy inequality $ab>1+\frac{3\pi}{2}$. So it seems to me that one should repeat all the steps of Hardy's proof.
As mentioned in the wiki entry, the function defined by $f(x)=\cases{{1\over \ln x},&$0<x\le 1/2$\cr \strut0, &$x=0$}$ is an example of a uniformly continuous function that is not Hölder continuous for any $\alpha>0$.
$f$ is continuous on $[0,1/2]$; and thus, since $[0,1/2]$ is compact, uniformly continuous on $[0,1/2]$.
But, $f$ is not Hölder continuous for any $\alpha>0$ at $x=0$. If it were, then there would exist positive $C$ and $\alpha$ such that $\bigl|0- {1\over \ln x}\bigr|\le C|x|^\alpha$ for all $0< x\le1/2$. But then, we would have $C|x|^\alpha |\ln x|\ge 1 $, which can't happen as $\lim\limits_{x\rightarrow0^+}C|x|^\alpha |\ln x|=0$.
According to the Wiki definition, $f$ is Hölder continuous for $\alpha=0$. That is, it is bounded. But one may extend $f$ to an unbounded, uniformly continuous function on $\Bbb R^+\cup\{0\}$ which is still not Hölder continuous at $x=0$.
Best Answer
($1$-dimensional) Brownian motion is almost surely continuous and nowhere Hölder continuous of order $\alpha$ if $\alpha > 1/2$. IIRC one can define random Fourier series that will be almost surely continuous but nowhere Hölder continuous for any $\alpha > 0$.
EDIT: OK, here's a construction. Note that $f$ is not Hölder continuous of order $\alpha$ at any point of $I = [0,1]$ if for every $C$ and every $x \in I$ there are $s,t \in I$ with $s \le x \le t$ and $|f(t)-f(s)|>C(t-s)^\alpha$.
I'll define $f(x) = \sum_{n=1}^\infty n^{-2} \sin(\pi g_n x)$, where $g_n$ is an increasing sequence of integers such that $2 g_n$ divides $g_{n+1}$. This series converges uniformly to a continuous function. Let $f_N(x)$ be the partial sum $\sum_{n=1}^N n^{-2} \sin(\pi g_n x)$. Note that $\max_{x \in [0,1]} |f_N'(x)| \le \sum_{n=1}^N n^{-2} \pi g_n \le B g_N$ for some constant $B$ (independent of $N$).
Now suppose $s = k/g_N$ and $t = (k+1/2)/g_N$ where $k \in \{0,1,\ldots,g_N-1\}$. We have $f(s) = f_{N-1}(s)$ and $f(t) = f_{N-1}(t) \pm N^{-2}$. Now $|f_{N-1}(t) - f_{N-1}(s)| \le B g_{N-1} (t-s) = B g_{N-1}/(2 g_N)$, so $|f(t) - f(s)| \ge N^{-2} - B g_{N-1}/(2 g_N)$. The same holds for $s = (k+1/2)/g_N$ and $t = (k+1)/g_N$. So let $g_n$ grow rapidly enough that $g_{n-1}/g_n = o(n^{-2})$ ($g_n = (3n)!$ will do, and also satisfies the requirement that $2g_n$ divides $g_{n+1}$). Then for every $\alpha > 0$, $(t-s)^\alpha = (2g_N)^{-\alpha} = o(N^{-2}) = o(|f(t) - f(s)|)$. Since for each $N$ the intervals $[s,t]$ cover all of $I$, we are done.