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 $0$.
If in your proof the value of $\delta$ doesn't depend on your choice of $x$, but uniquely on $\varepsilon$, then you could have fixed $x$ after fixing $\varepsilon$ and determining $\delta$, therefore proving absolute continuity of the function, hence you can basically use the same proof, except for fixing $x$ and $y$ after determining $\delta$.
Best Answer
We say that $a_n$ and $b_n$ are equivalent if $a_n-b_n \to 0$. If $f$ is uniformly continuous then maps equivalent sequences to equivalent sequences.
1) Consider the sequences $a_n =(1/n)$ and $b_n=(1/2n)$ both are equivalent in $(0,1]$. But $f(a_n)$, $f(b_n)$ are not equivalent. So $f$ is not uniformly continuous in $(0,1]$.
[Alternatively you can show that if $f$ is uniformly continuous then maps Cauchy sequences to Cauchy sequences, and in particular for this case, $(1/n)$ is a Cauchy sequence in $(0,1]$, but $f(1/n)$ is not a Cauchy sequence].
2) You can show that uniformly continuous using $\varepsilon$-$\delta$ definition is logically equivalent to say that $f$ maps equivalent sequences to equivalent sequences, i.e., $f(a_n)-f(b_n)\to 0$ whenever $a_n -b_n \to 0$. Using this second definition:
Given $(a_n), (b_n) \subset [1, \infty)$ and $a_n-b_n \to 0$. We shall show that $f(a_n)-f(b_n) \to 0$. Let $\varepsilon>0$ be arbitrary and choose $n_0>0$ such that $|a_n-b_n|<\varepsilon$ for all $n\ge n_0$. Thus
$$|f(a_n)-f(b_n)|= \bigg|\frac{1}{a_n}-\frac{1}{b_n} \bigg|=\frac{|a_n-b_n|}{a_n b_n}\le |a_n-b_n|< \varepsilon$$
This is possible since $1\le a_n, b_n$. Hence $f(a_n)-f(b_n) \to 0$.
3) Consider the sequences $a_n = n$ and $b_n = n+1/n$ both are equivalent sequences in $[0,\infty)$. But $f(b_n)=n^2+2+1/n^2=f(a_n)+2+1/n^2$, so $f(b_n)-f(a_n)\ge2$. Thus we can conclude that is not uniformly continuous.
4) Since $f$ is continuous on $[0,1]$ then must be uniformly continuous (why?). Let $(0,1) \hookrightarrow [0,1]$. We claim that is uniformly continuous, let $x_n-y_n\to 0$ and $(x_n),(y_n) \subset (0,1)$, so $i(x_n)-i(y_n) = x_n -y_n \to 0$, since the sequences are arbitrary it holds for all equivalent sequences and so $i$ is uniformly continuous.
We claim that $f\circ i$ is uniformly continuous. We shall show that maps equivalent sequences to equivalent sequences. Since $i(x_n)-i(y_n)\to 0$ and $f$ is uniformly continuous then $ f(i(x_n))-f(i(y_n))=(f\circ i)(x_n)-(f \circ i)(y_n) \to 0$.
Hence $f\circ i$ is uniformly continuous. But $f\circ i=f \restriction _{(0,1)} $ which is just the square root function define on $(0,1)$.
5) Suppose that $(a_n),(b_n) \subset [1,\infty)$ and $a_n-b_n \to 0$. We shall show that $a_n^{1/2}-b_n^{1/2} \to 0$. Given $\varepsilon>0$, choose $n_0$ such that $|a_n-b_n|< \varepsilon$ for all $n\ge n_0$. Thus
$$|f(a_n)-f(b_n)|=|\sqrt{a_n}-\sqrt{b_n}|= \bigg|\frac{a_n-b_n}{\sqrt{a_n}+\sqrt{b_n}} \bigg|=\frac{|a_n-b_n|}{\sqrt{a_n}+\sqrt{b_n}}\le \frac{|a_n-b_n|}{2}< \varepsilon$$
Since $1\le a_n, b_n$.