Suppose $f$ is continuous on [0, 1]. Must there be a nondegenerate closed subinterval $[a, b]$ of [0, 1] for which the restriction of $f$ to $[a, b]$ is of bounded variation?
$\mathbf{My\ attempt}:$ Suppose for all $[a,b]\subseteq [0,1]$ with $a\neq b$, we have $f$ restricted to $[a,b]$ is not of a bounded variation, $i.e$ for all $n \in \mathbb{Z^+}$ there exists a partition $\mathcal{P}_{{n}}$ such that $V(f,\mathcal{P_n})\geq n$.
I am not sure if there would be some examples shows that is not true, I am using a contradiction here to conclude that there must be a nondegenerate interval of $[0,1$ such that $f$ restricted to this interval is of a bounded variation. I will appreciate it any help with that. Thank you.
Hint: it is well known that functions of bounded variation are of the form $f = g-h$ with $g,h$ monotone. In particular, $f$ ought to be differentiable almost everywhere on $[0,1]$. However, functions such as the Weierstraß function (restricted to $[0,1]$) are continuous but differentiable nowhere. In particular it is not differentiable in any proper subinterval and so by the former, it is not of bounded variation there either.
Edit: the original question was modified, and this does not answer the content of the post: what follows is a construction of a continuous function $f : [0,1] \to \mathbb{R}$ which is not of bounded variation in any subinterval $[0,c]$ with $c \leq 1$, but for which $V_c^1f < +\infty$.
First, let's construct a function $[0,1] \to \mathbb{R}$ which is not of bounded variation.
Consider $f_N : \left[\frac{1}{N+1},\frac{1}{N}\right]\to \mathbb{R}$ with interpolates linearly the points: $$ (1/(N+1),0),\left(\frac{1/(N+1)+1/N}{2},1/N\right) \text{ and } (1/N,0) $$
A drawing should illustrate the idea: we are making spikes whose height decreases with $N$, and so does the length of the considered interval. At the endpoint of the interval $f_N$ is zero, and in the middle point it attains its maximum value which is $1/N$. Moreover, $f_N$ is piecewise monotone. Now the functions $(f_N)_{N\geq 1}$ coincide at $\{1/N : N \in \mathbb{N}\}$ and so the function
$$ f(x) = f_N(x) \text{, if } x \in \left[\frac{1}{N+1},\frac{1}{N}\right] $$
is continuous on $(0,1]$. Technicalities aside, we are just gluing the spikes together and justifying that the resulting function is indeed contiunous. Finally, we would like to extend $f$ to $[0,1]$. A necessary condition is for $f(0)$ to be zero, since we already have $f(1/N) = 0$ and $1/N \to 0$. So let's check that
$$ f(x) = \cases{f(x) \ x \neq 0 \\ 0 \text{ otherwise}} $$
is continuous on $[0,1]$. Take $x_n \to 0$ and $\varepsilon > 0$. Since $x_n$ goes to zero, at some point $n_0$ we have that $|x_n| < \varepsilon$ if $n \geq n_0$. Now, if $n \geq n_0$, each $x_n$ is either zero, lies on some interval $\left[\frac{1}{M_n+1},\frac{1}{M_n}\right]$ with $M_n > 1/\varepsilon$, and so by the definition of each $f_N$ then, we have that $|f(x_n)| = f(x_n) \leq 1/M_n < \varepsilon$. In any case, we have that $|f(x_n)| < \varepsilon$ if $n \geq n_0$, which proves that $f$ is continuous at zero.
So far we have only defined $f$. We still ought to show that it is not of bounded variation, despite being continuous. Intuitively this is because the spikes add each a variation of no less that $1/N$ and the harmonic series diverges. Concretely, if $f$ were of bounded variation, we would have that for each $N \geq 1$,
$$ V_0^1f = V_0^{1/(N+1)}f + \sum_{i=1}^NV_{1/(i+1)}^{1/i}f = \dots \\ \dots = V_0^{1/(N+1)}f + \sum_{i=1}^NV_{1/(i+1)}^{1/i}f_i \geq V_0^{1/(N+1)}f + \sum_{i=1}^N\frac{1}{i} \geq \sum_{i=1}^N\frac{1}{i}. $$
Taking limit when $i \to \infty$ we have $V_0^1f = +\infty$. Moreover, $f$ is not of bounded variation in any interval of the form $[0,c]$: if it were so, by refining partitions and taking limits we would have that
$$ V_0^1f = V_0^cf + V_c^1f, $$
and since $f|_{[c,1]}$ is of bounded variation (we are interpolating linearly a finite amount of times), then $V_c^1 < + \infty$ which would imply that $V_0^1f < + \infty$, a contradiction.