Setting $\varepsilon$ to something doesn't make sense. You need to take $\varepsilon$ to be given, and find a value of $\delta$ that's small enough.
Continuity should not say $\exists c\in(0,1]$ etc., where $c$ is in the role you put it in. Rather, continuity at the point $c$ should be defined by what comes after that.
Uniform continuity says
$$
\forall\varepsilon>0\ \exists\delta>0\ \forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right).
$$
Lack of uniform continuity is the negation of that:
$$
\text{Not }\forall\varepsilon>0\ \exists\delta>0\ \forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right). \tag 1
$$
The way to negate $\forall\varepsilon>0\ \cdots\cdots$ to by a de-Morganesque law that says $\left(\text{not }\forall\varepsilon>0\ \cdots\cdots\right)$ is the same as $(\exists\varepsilon>0\ \text{not }\cdots\cdots)$, and similarly when "not" moves past $\forall$, then that transforms to $\exists$. So $(1)$ becomes
$$
\exists\varepsilon>0\text{ not }\exists\delta>0\ \forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 2
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\text{ not }\forall x\in(0,1]\ \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 3
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\text{ not } \forall y\in(0,1]\ \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 4
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\ \exists y\in(0,1]\text{ not } \left(\text{if }|x-y|<\delta\text{ then }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 5
$$
and that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\ \exists y\in(0,1] \left(|x-y|<\delta\text{ and not }\left|\frac1x-\frac1y\right|<\varepsilon\right) \tag 6
$$
and finally that becomes
$$
\exists\varepsilon>0\ \forall\delta>0\ \exists x\in(0,1]\ \exists y\in(0,1] \left(|x-y|<\delta\text{ and }\left|\frac1x-\frac1y\right|\ge\varepsilon\right). \tag 7
$$
To show that such a value of $\varepsilon$ exists, it is enough to show that $\varepsilon=1$ will serve. You need to find $x$ and $y$ closer to each other than $\delta$ but having reciprocals differing by more than $1$. It is enough to make both $x$ and $y$ smaller than $\delta$ and then exploit the fact that there's a vertical asymptote at $0$ to make $x$ and $y$ far apart, by pushing one of them closer to $0$.
@Craig I'll answer here as the comments section is quite cramped.
"Do they make the function monotonous just so it doesn't oscillate wildly?" Exactly. Notice that if you are adding up $\sum_{i=1}^m |f(d_i)-f(c_i)|$ for
nonoverlapping intervals in $[0,W]$ then
$$\sum_{i=1}^m |f(d_i)-f(c_i)| =\sum_{i=1}^m [ f(d_i)-f(c_i)] \leq f(W)-f(0).$$
So all you need to control these for a monotonic increasing continuous function is to make sure that $f(W)-f(0)$ is small.
"Uniform continuity?" Well if $f$ is continuous on $[0,1]$ it is uniformly continuous there, if you feel you need it.
So now the idea of the proof. Start with (as always) "Let $\epsilon>0$.
First we set up the wall $W$. Choose $W>0$ so that $f(W)-f(0)< [small_1]$.
Now $f$ is absolutely continuous on $[W,1]$ (the right side of the wall) so we can select $\delta_r>0$ so that
$$\sum_{i=1}^m |f(d_i)-f(c_i)| \leq [small_2] $$
whenever these intervals are nonoverlapping subintervals of $[W,1]$ with total length less than $\delta_r$.
Now $f$ is continuous at $W$ (the wall) so there is a $\delta_W$ with
$$f(W+\delta_W) - f(W-\delta_W) < [small_3] .$$
Now consider any finite sequence of intervals $\{[c_i,d_i]\}$ of $[0,1]$ with total length less than $\delta_?$ [still to be determined as this isn't the actual proof just our heuristics to find a proof].
That sum has three parts: the pieces in $[0,W]$ on the left side of the wall
(ii) the pieces in $[W,1]$ on the right side of the wall and (iii) maybe one solitary piece straddling the wall. Give an $\epsilon/3$ to each of these possibilities and we are done and can write up a tidy $\epsilon$, $\delta$ proof.
P.S. Don't leave the problem without a counterexample showing that the hypotheses that you used (and that were stated in the problem) cannot be dropped). This is proper mathematical etiquette: think of it as saying "please" and "thankyou" in social situations. Add "counterexample" to your
solutions whether the problem asks it or not. Never be impolite and just answer the problem, rushing away for more important things. Obviously continuity cannot be dropped. Find an example with the oscillatory behaviour that you mentioned above happening at the left endpoint. The example will have to have unbounded variation on $[0,1]$ even though it is absolutely continuous (and hence BV) on $[W,1]$ for all $W>0$.
Best Answer
I think I managed to figure it out.
Remark: this then also means that the function we could define from the $d$-s is continous, and so on ad infinitum.