Your guess is indeed right. Following a similar idea gives you the Takagi or blancmange function.
It is even quasi-Lipschitz (it has a modulus of continuity $\omega(t)=ct(|\log(t)|+1)$ for a suitable constant $c>0$), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus $W^{1,p} _ {loc}$ for no $p\geq1$.
Rmk 1. The above example is for dimension 1: but of course it holds in any dimension a fortiori.
Rmk 2. To get an example with a more classical flavor, actually a Weierstrass function, replace $s(x)$ with $\cos(x)$. I'd say that the resulting Fourier series defines a function with the same features, by the same reasons (the function $\cos(x)$ works better than $\sin(x),$ in view of point 2 below.)
Rmk 3. Once you know that the Weierstrass function $f(x):=\sum_{k=0}^\infty 2^{-k}\cos(2^k x)$ is nowhere differentiable, you also have that it is BV on no open interval, for BV on an interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval.
Details.
To prove that the Takagi function $f(x)$ admits the above modulus of continuity, recall that that $f$ is characterized as the fixed point of the affine contraction $T:C_b(\mathbb{R})\to C_b(\mathbb{R})$ such that $(Tf)(x)=\frac{1}{2}f(2x)+s(x),$ for all $x\in\mathbb{R}$, where $s(x)$ is the distance function from the integers (a zig-zag piecewise 1-periodic function). Just find a $c$ such that the subset of $C_b(\mathbb{R})$ of functions that admit $\omega$ as modulus of continuity is a $T$-invariant set. The latter subset is obviously closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle).
Proving that $f$ is not of bounded variation on $[0,1]$ (hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the partial sum $f_n$ of the series defining $f$. Let
$$f_n(x):=\sum_{k=0}^{n-1}\, 2^{-k} s(2^k x).$$
First note that the derivative of $f_n$ only takes integer values, which of course come as a result of the sum of $n$ terms $\pm 1$ (with all the $2^n$ possible signs). In particular, for any $n\in\mathbb{N}$ the function $f_{2n}$ has ${2n \choose n} $ flat intervals of lenght $2^{-2n}$ within the unit interval $I$, and has derivative larger than $2$ in absolute value elsewhere in $I$. Thus, for the subsequent odd index $2n+1,$ the function $f_{2n+1}$ has
${2n \choose n}$ local maxima in $I$ (located in the mid-points of the above intervals). Moreover, passing to $f_{2n+1}$ each maximum point contributes to the increment of the total variation with $2^{-2n}$, while the total variation remains unchanged passing from $2n+1$ to the next even index $2n+2$. The conclusion is that, for any $n$, the total variation of $f_n$ on $I$ is
$$V(f_n;I)=\sum_{0\leq k < n/2}{2k\choose k}2^{-2k} =O\big(\sqrt{n}\big),$$
since by the classical asymptotics for the central binomial coefficient, ${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\, k\to\infty.$ So actually $V(f_n;I)$ diverges. Yet this would not be sufficient to conclude that $V(f,I)=\infty,$ as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation on a given subdivision $P:=\{t_0 < \dots < t_r \}$
$$V(f_n; P\, )=\sum_{i=0}^{r-1}\, \big|f_n(t_{i+1})-f(t_i)\big|$$
does of course pass to the limit under even pointwise convergence. Now the point is that, for the binary subdivision $P_m:=\{ k2^{-m} \, : \, 0 \le k \le 2^m \},$ we have $V(f_n;I)=V(f_n;P_m)$ as soon as $n \geq m$. So for all $m$ letting $n\to\infty$
$$V(f;P_m)=\lim_{n\to\infty }V(f_n;P_m)=V(f_m;P_m)$$ and
$$V(f;I)=\sup_{m\in\mathbb{N}}V(f;P_m)=\infty,$$
as we wished to show.
This is a good question, given the way calculus is currently taught, which for me says more about the sad state of math education, rather than the material itself. All calculus textbooks and teachers claim that they are trying to teach what calculus is and how to use it. However, in the end most exams test mostly for the students' ability to turn a word problem into a formula and find the symbolic derivative for that formula. So it is not surprising that virtually all students and not a few teachers believe that calculus means symbolic differentiation and integration.
My view is almost exactly the opposite. I would like to see symbolic manipulation banished from, say, the first semester of calculus. Instead, I would like to see the first semester focused purely on what the derivative and definite integral (not the indefinite integral) are and what they are useful for. If you're not sure how this is possible without all the rules of differentiation and antidifferentiation, I suggest you take a look at the infamous "Harvard Calculus" textbook by Hughes-Hallett et al. This for me and despite all the furor it created is by far the best modern calculus textbook out there, because it actually tries to teach students calculus as a useful tool rather than a set of mysterious rules that miraculously solve a canned set of problems.
I also dislike introducing the definition of a derivative using standard mathematical terminology such as "limit" and notation such as $h\rightarrow 0$. Another achievement of the Harvard Calculus book was to write a math textbook in plain English. Of course, this led to severe criticism that it was too "warm and fuzzy", but I totally disagree.
Perhaps the most important insight that the Harvard Calculus team had was that the key reason students don't understand calculus is because they don't really know what a function is. Most students believe a function is a formula and nothing more. I now tell my students to forget everything they were ever told about functions and tell them just to remember that a function is a box, where if you feed it an input (in calculus it will be a single number), it will spit out an output (in calculus it will be a single number).
Finally, (I could write on this topic for a long time. If for some reason you want to read me, just google my name with "calculus") I dislike the word "derivative", which provides no hint of what a derivative is. My suggested replacement name is "sensitivity". The derivative measures the sensitivity of a function. In particular, it measures how sensitive the output is to small changes in the input. It is given by the ratio, where the denominator is the change in the input and the numerator is the induced change in the output. With this definition, it is not hard to show students why knowing the derivative can be very useful in many different contexts.
Defining the definite integral is even easier. With these definitions, explaining what the Fundamental Theorem of Calculus is and why you need it is also easy.
Only after I have made sure that students really understand what functions, derivatives, and definite integrals are would I broach the subject of symbolic computation. What everybody should try to remember is that symbolic computation is only one and not necessarily the most important tool in the discipline of calculus, which itself is also merely a useful mathematical tool.
ADDED: What I think most mathematicians overlook is how large a conceptual leap it is to start studying functions (which is really a process) as mathematical objects, rather than just numbers. Until you give this its due respect and take the time to guide your students carefully through this conceptual leap, your students will never really appreciate how powerful calculus really is.
ADDED: I see that the function $\theta\mapsto \sin\theta$ is being mentioned. I would like to point out a simple question that very few calculus students and even teachers can answer correctly: Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians. In my department we audition all candidates for teaching calculus and often ask this question. So many people, including some with Ph.D.'s from good schools, couldn't answer this properly that I even tried it on a few really famous mathematicians. Again, the difficulty we all have with this question is for me a sign of how badly we ourselves learn calculus. Note, however, that if you use the definitions of function and derivative I give above, the answer is rather easy.
Best Answer
One reason is this: if $f$ is in $H^1({\mathbb R^n})$, you have $\int_{\mathbb R^n}|f(x+h)-f(x)|^2 dx\le C|h|^2$ for all $h \in{\mathbb R^n}$ . Now in the case of $f:=\chi_E$ the integral is just the $L^1$ distance, $\|\chi_E - \chi_{E-h}\|_1$. By the triangular inequality, for any positive integer m and any h as above one gets
$\|\chi_E -\chi_{E-h}\|_1 \le$
$\sum_{j=0}^{m-1}\|\chi_{E - \frac{j}{m} h } - \chi_{E-\frac {j+1}{m}h}\|_1 \le$
$Cm|h/m|^2=C|h|^2/m$,
whence $\|\chi_E - \chi_{E-h}\|_1=0$ for all $h$. This is impossible since for $|h|\to\infty$ it converges to 2meas(E).