Open intervals $(a,b), a < b$ form a base for the topology of $\mathbb{R}$.
What is $f^{-1}[(a,b)] = \left\{x \in \mathbb{R}: x^2 \in (a,b) \right\}$?
If $b \le 0$, then no square is in $(a,b)$ so then $f^{-1}[(a,b)] = \emptyset$, which is open. So assume $b > 0$. Then $x^2 < b$ iff $x \in (-\sqrt{b},\sqrt{b})$.
If $a < 0$, the $a$ does not impose an extra condition, as all $x^2 \ge 0 > a$ in that case, so
$f^{-1}[(a,b)] = (-\sqrt{b},\sqrt{b})$ if $b > 0, a < 0$, which is an open interval in $\mathbb{R}$ so open.
Otherwise we also need $x^2 > a \ge 0$, so $x < -\sqrt{a}$ or $x > \sqrt{a}$ respectively.
So then $f^{-1}[(a,b)] = (-\sqrt{b}, -\sqrt{a}) \cup (\sqrt{a},\sqrt{b})$ if $b > a \ge 0$, which is open as the union of two open intervals.
This covers all cases, so $f^{-1}[(a,b)]$ is open for all intervals.
Now if $O$ is open, we can write $O = \bigcup_{i \in I} (a_i,b_i)$, for some family of open intervals $(a_i,b_i), i \in I$, as the intervals form a base. But then
$$f^{-1}[O] = f^{-1}[\bigcup_{i \in I} (a_i,b_i)] = \bigcup_{i \in I} f^{-1}[(a_i,b_i)]$$
by standard properties of $f^{-1}$ and the last set is open as unions of open sets are open, and we have shown that the inverse images of the base sets are open.
You basically have it. You just need to be a little more careful. You are claiming that for $\epsilon=1$, and for every $\delta>0$, there are $x,y\in (0,1)$ such that$|x-y|<\delta$ and $|f(x)-f(y)|>1.$ You may assume without loss of generality that $\delta<1/2$ because it the claim is true for all such $\delta,$ it will be true for any value of $\delta$ larger than $1/2.$
(Remember, all you need to do is find two numbers in $(0,1)$ whose difference is less than $\delta$ in absolute value. The ones that work for $\delta<1/2$ will also work for $\textit{any}\ \delta\ge 1/2.$ Example: suppose we have $\delta=15$ and you can find $x,y$ such that $|x-y|<1/2$ and $|f(x)-f(y)|>1.$ Then, the $x,y$ work for $\textit{both}$ values of $\delta$ simultaneously because if $|x-y|<1/2$ it is also $<15$).
Now, $|f(x)-f(y)|=\left|\frac{x-y}{xy}\right|$ and we want to choose $x$ and $y$ so that $|x-y|<\delta$ but $\left|\frac{x-y}{xy}\right|>1$, so take $x=\delta$ and $y=2\delta.$ Then, $x$ and $y$ are actually in $(0,1)$ and $\left|\frac{x-y}{xy}\right|=\frac{2}{\delta}>1$, and you are done.
It may be easier to do it with sequences: with $\epsilon=1/2,$ take $\delta_n=1/n$ and find sequences $(x_n)$ and $(y_n)$ such that $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|>1/2.$ Choose $x_n=1/n$ and $y_n=1/n+1$ and check that this assignment works.
Best Answer
This is one of those cases where maybe it's better --and more instructive--to prove the general case first. Then, adapting the proof to your specific function is easy.
So, suppose $v,w\in \mathbb F^{n}$. Then,
$w=v+(w-v)$ so $\Vert w\Vert \leq \Vert v\Vert +\Vert w-v\Vert \Rightarrow\Vert w\Vert-\Vert v\Vert\leq \Vert w-v\Vert$.
Interchanging the roles of $v$ and $w$ gives $\Vert v\Vert-\Vert w\Vert\leq \Vert v-w\Vert=\Vert w-v\Vert$
so in fact
$\left | \Vert w\Vert-\Vert v\Vert\ \right |\leq \Vert w-v\Vert$, which proves that $\Vert \cdot\Vert $ is continuous (with $\delta =\epsilon)$.