First, uniform continuity is about an interval, not just a point. You can say f(x) is continuous on the interval $(a,b)$ means: $\lim_{x \rightarrow c} f(x) = f(c)$ for all $c$ in $(a,b)$. Then you have to say what you mean by a limit, and you get the old $\varepsilon$-$\delta$ statement: for every $\varepsilon$ there exists a $\delta$ such that $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon$.
Now what they never tell you is that $\delta$ depends on $x$. For some functions $\delta$ has to get smaller and smaller to get $|f(x) -f(c)| < \varepsilon$. An example of this is the function $f(x) = 1/x$ on $(0,1)$. It's continuous there everywhere, but as $x$ approaches $0$ the function gets steeper and steeper. That means you have to take your $x$ and $c$ closer and closer together -- i.e $\delta$ smaller and smaller, to get $|f(x) -f(c)| < \varepsilon$.
What uniform continuity means is that the $\delta$ does NOT depend on x. You can find a single $\delta$ that depends only on $\epsilon$ for the entire interval.
Intuitively uniform continuity means your function can't get infinitely steeper on $(a,b)$ the way $1/x$ does on $(0,1)$. Steepness doesn't only mean that $f$ may go off to infinity in your interval. It could also oscillate around in some unpleasant way. Look for example at $f(x) = \sin(1/x)$ on $(0,1)$.
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 $0$.
Best Answer
Set $I=(0,a]$ and $J=[a,\infty)$. Let $\epsilon>0$.
Choose $\delta_1>0$ such that $|f(x)-f(y)|<\epsilon/2$ for all $x\in I$ and $y\in I$ with $|x-y|<\delta_1$.
Choose $\delta_2>0$ such that $|f(x)-f(y)|<\epsilon/2$ for all $x\in J$ and $y\in J$ with $|x-y|<\delta_2$.
Let $\delta=\min\{\delta_1,\delta_2\}$.
Then if $|x-y|<\delta$:
If $x$ and $y$ are both in $I$ or both in $J$, $|f(x)-f(y)|<\epsilon/2<\epsilon$.
If $x\in I$ and $y\in J$, or $x\in J$ and $y\in I$
$$ |f(x)-f(y)|\le |f(x)-f(a)|+|f(a)-f(y)|<\epsilon/2 +\epsilon/2=\epsilon. $$