I wanted to verify my proof as it seems correct but it does not look convincing.
I first chose $\delta > 0$ and $\epsilon > 0$ to be fixed and arbitary and consider $|x-y| = a$. Without loss of generality I then say $0<x<y\leq 1$.
I then set $y = \sqrt(\frac{\epsilon}{a})$ and $\epsilon \leq a < \delta$.
Thereafter I go onto the proof and say
$|x-y| = a < \delta$ holds true and we can evaluate $|f(x)-f(y)| = |\frac{1}{x}-\frac{1}{y}| = \frac{a}{xy} \geq \frac{a}{y^2} = \epsilon.$ which mean $|f(x)-f(y)| \geq \epsilon$.
This thereby concludes the proof and shows $f(x)$ is not uniformly continuous over the interval.
Now I problem that I am facing is that I am not sure how many assumptions I am allowed to make and work out the inequalities and if my above work has some flaw in it that I am missing.
Unfortunately, this proof is not correct. You mentioned that it didn't look convincing, and you had good intuition! Now lets try to unpack why this didn't feel convincing.
If you think of a proof like a template for a story, then the template should only be able to generate a story given the correct inputs. And so if you have a proof that a function is not uniformly continuous, which can be rewritten with a function which is uniformly continuous, then there is clearly a mistake in your reasoning, since through renaming alone you would then be able to derive a contradiction. This is precisely the case with this proof. To see why, let's take
$$f:(0,1]\to\mathbb{R} $$
Sending $x\mapsto x$. This is clearly uniformly continuous, in fact it is an isometry.
But now choose $a<\delta$, and assume $|x-y|= a$, and w.l.o.g assume $x\leq y\leq 1$. Then let
$$\varepsilon = a $$
Then
We have $|f(x)-f(y)|\geq \varepsilon$
So $f(x)$ is not uniformly continuous.
Its difficult to isolate a single mistake in this proof, but I suspect it came down to the negation of the uniform continuity. You proved
$$\exists_{\delta}\exists_{a<\delta}\exists_{\varepsilon}\forall_{x,y}[|x-y|=a\implies |f(x)-f(y)|\geq \varepsilon$$
The correct negation is
$$\exists_{\varepsilon}\forall_{\delta}\exists_{x,y}[|x-y|<\delta \land |f(x)-f(y)|\geq \varepsilon]$$
If you do not understand why this should be the negation, I suggest asking a separate question focusing only on this