Heh. Yes, what's below is so an answer to the question. He didn't say he didn't want to use that theorem, he said he wanted an elementary proof. We give a totally elementary proof of the relevant half of the theorem:
Theorem Suppose $f:[a,b]\to\Bbb R$ is nondecreasing and satisfies $f'=0$ almost everywhere. Then $L(f)=b-a+f(b)-f(a)$.
Note In fact a monotone function must be differentiable almost everywhere, but we're not using that here. To be quite explicit the hypothesis is this: There exists a set $E$ of measure zero such that if $x\in[a,b]\setminus E$ then $f'(x)$ exists and equals zero.
The idea of the proof is simple: We cover most of the set where $f'=0$ by intervals where $f$ is close to constant. The sum of the lengths of those intervals is close to $b-a$ since $f'=0$ almost everywhere. And since $f$ is close to flat on those intervals, the total variation of $f$ on those intervals is small, so the variation of $f$ on the rest of $[a,b]$ is close to $f(b)-f(a)$. This is so far really what's going on in various arguments we've seen for the Cantor function. But since now the set where $f'=0$ need not be open or closed and we don't want to use measure theory there are some technicalities:
Proof: Say $\gamma(x)=(x,f(x))$. If $s<t$ then $||\gamma(s)-\gamma(t)||\le t-s+f(t)-f(s)$ by the triangle inequality; hence $$L(f)\le b-a+f(b)-f(a).$$
For the other inequality, let $\epsilon>0$. Suppose $E$ has measure zero and $f'=0$ on $[a,b]\setminus E$. Choose intervals $I_j=(a_j,b_j)$ with $$E\cup\{a,b\}\subset\bigcup_{j=1}^\infty I_j$$and $$\sum(b_j-a_j)<\epsilon.$$Wew can assume the $I_j$ are disjoint (If necessary, replace $(I_j)$ by the connected components of $\bigcup I_j$.)
Let $K=[a,b]\setminus\bigcup I_j$. Note that $K$ is a compact subset of $(a,b)$. If $x\in K$ then $f'(x)=0$, hence there exists $\delta_x>0$ such that $$f(x+\delta_x)-f(x-\delta_x)<2\epsilon\delta_x.$$Since $K$ is compact we have $K\subset\bigcup_{k=1}^NJ_k$ where $$J_k=(\alpha_k,\beta_k)$$and $$f(\beta_k)-f(\alpha_k)<\epsilon(\beta_k-\alpha_k).$$We may assume the $J_k$ are disjoint, since if two of them should intersect then their union is $(\alpha,\beta)$ with $f(\beta)-f(\alpha)<\epsilon(\beta-\alpha)$.
Since we have finitely many disjoint intervals, we may relabel them from left to right: $$\beta_k\le\alpha_{k+1}.$$
Note that if $s<t$ then $$||\gamma(s)-\gamma(t)||\ge t-s$$and $$||\gamma(s)-\gamma(t)||\ge f(t)-f(s).$$So considering the partition $$
a\le\alpha_1<\beta_1\le\alpha_2\dots<\beta_N\le b$$shows that $$L(f)\ge
f(\alpha_1)-f(a)+\sum(\beta_k-\alpha_k)+\sum(f(\alpha_{k+1})-f(\beta_k))+f(b)-f(\beta_N).\quad(1).$$
Now $$f(b)-f(a)=
f(\alpha_1)-f(a)+\sum(f(\beta_k)-f(\alpha_k))+\sum(f(\alpha_{k+1})-f(\beta_k))+f(b)-f(\beta_N)$$and $$\sum(f(\beta_k)-f(\alpha_k))
<\epsilon\sum(\beta_k-\alpha_k)\le\epsilon(b-a),$$so $$f(\alpha_1)-f(a)+\sum(f(\alpha_{k+1})-f(\beta_k))+f(b)-f(\beta_N)>f(b)-f(a)-\epsilon(b-a)\quad(2).$$
Since $[a,b]\setminus\bigcup I_j=K\subset\bigcup J_k$ we have $$[a,b]\subset\bigcup I_j\cup\bigcup J_k.$$
Now, it's easy to see by induction on $n$ that if $[a,b]$ is covered by $n$ open intervals then the sum of the lenghts of the intervals must be at least $b-a$. (Say $(\alpha,\beta)$ is one of the $n$ intervals and $b\in(\alpha,\beta)$; then $[a,\alpha]$ is covered by the $n-1$ other intervals...) So, since $I_j$ and $J_k$ are open and $[a,b]$ is compact the previous display implies that $$b-a\le\sum(b_j-a_j)+\sum(\beta_k-\alpha_k).$$Hence$$\sum(\beta_k-\alpha_k)\ge b-a-\epsilon\quad(3).$$
Combining (1), (2), and (3) shows that $$L(f)\ge b-a+f(b)-f(a)-\epsilon(b-a+1);$$hence $L(f)\ge b-a+f(b)-f(a)$.
