I would like to prove that a function defined $[a,b]$ of bounded variation is bounded by contradiction : suppose that $f$ is of bounded variation while note bounded and come up with a contradiction. Here is how I have started.
Suppose that $f$ is not bounded : $\forall M\in\mathbb{R}_{+},\exists x\in[a,b] : \lvert f(x)\rvert>M$
Consider the sequence $u_n = n$, then for each $n\in\mathbb{N}$ we can find an $x_n$ such that $\lvert f(x_n)\rvert> n$.
And then I don't see how to pursue (if this is the good starting point). I would like to consider a partition of $[a,b]$ including these points but I am not sure if it is the right way to prove this.
I would like your help please !
Edit (Thanks to jro)
Suppose $f$ is not bounded on $[a,b]$ but of bounded variation, that is $V(f) <M$ for some $M\in\mathbb{R}_{+}$.
For each $n\in\mathbb{N}$ we can find $x_n\in[a,b]$ such that $\lvert f(x_n)\rvert>n + \lvert f(x_0)\rvert$.
Now we will show that we can find a partition of $[a,b]$ such that the total variation is not bounded.
We can find $n\in\mathbb{N}$ such that $n>M$. Thus we have by using what we have above :
$$ \lvert f(x_n) - f(x_0)\rvert\geq \lvert \lvert f(x_n)\rvert - \lvert f(x_0)\rvert\rvert>n>M $$
Then for any partition $(x_i)_{i=0}^{N}$ of $[a,b]$ including $x_n$ with $0\leq n\leq N$ we have
$$ \sum_{i=0}^{N-1}\lvert f(x_{i+1}) - f(x_i)\rvert=\sum_{i=0}^{n-1}\lvert f(x_{i+1}) - f(x_i)\rvert + \sum_{i=n}^{N-1}\lvert f(x_{i+1}) - f(x_i)\rvert\geq \sum_{i=0}^{n-1}\lvert f(x_{i+1}) - f(x_i)\rvert $$
It follows that $$ \sum_{i=0}^{n-1}\lvert f(x_{i+1}) - f(x_i)\rvert\geq\lvert\sum_{i=0}^{n-1}[f(x_{i+1}) - f(x_i)]\rvert = \lvert f(x_n) - f(x_0)\rvert > n > M $$
So we have
$$ V(f) = \sup_{\sigma\in S}\sum_{i=0}^{N-1}\lvert f(t_{i+1}) - f(t_i)\rvert\geq \sum_{i=0}^{N-1}\lvert f(x_{i+1}) - f(x_i)\rvert > M $$
where the supremum is taking over all the partition, $\sigma$, of $[a,b]$.
Which is not possible by assumption. Thus, we have shown that being of bounded variation for a function on $[a,b]$ is a sufficient condition to be bounded.