Prove that when x approaches to 1, function 1/(x-1) doesn't have limit

You have all the right ingredients and all the right tools, but I can see from your write-up alone that you are finding the four nested quantifiers difficult. And it is difficult to properly wrap your head around. It might help if you start with a skeleton for your proof, that you can fill in.

Because your statement starts with $\forall L \in \Bbb{R}$, a good sentence to begin your proof would be

Suppose $L \in \Bbb{R}$.

The next quantifier is $\exists \varepsilon > 0$. So, at some point after this first sentence, we will need

Let $\varepsilon = \underline{\hspace{20pt}}$ [a function of $L$, taking into account the definition of the function $f$].

The specific value for this $\varepsilon$ is for us to decide. This should be based on our scratch work. The next quantifier is $\forall \delta > 0$, so once again, we will need the following:

Suppose $\delta \in \Bbb{R}$.

Based on our final quantifier, we then want:

Let $x = \underline{\hspace{20pt}}$ [a function of $L$ and $\delta$, taking into account $f$ and possibly our choice of $\varepsilon$].

That's when we can start proving the predicate at the end of the quantifiers: that $x \in \operatorname{dom} f$, that $|x - 1| < \delta$, and that $|\frac{1}{x - 1} - L| \ge \varepsilon$.

Now, let's fill in the gaps. From your proof, $\varepsilon > 0$ seemed to be arbitrary. All you needed was some $n \ge L + \varepsilon$. So, let's just pick $\varepsilon = 1$. And why not! You should be able to find some $n \ge L + 1$.

Next, we need to figure out our $x$. You chose $x = \frac{1}{n} + 1$, which is potentially sensible, but you should really define your $n$ first. How was $n$ defined in your proof? It was defined to satisfy two properties:

1. $n \ge L + \varepsilon = L + 1$, and
2. $\frac{1}{n} < \delta$.

Both of these are sensible conditions based on the Archimedean principle. Let's define: $$n = \max\left\{\lceil L \rceil + 1, \left\lceil \frac{1}{\delta}\right\rceil + 1\right\},$$ where $\lceil \cdot \rceil$ is the ceiling function (which only exists due to the Archimedean property!). Then, it makes sense to define $x = 1 + \frac{1}{n}$. With this in mind, I would write the proof as follows:

Suppose $L \in \Bbb{R}$. Let $\varepsilon = 1$, and suppose $\delta \in \Bbb{R}$. Define $$n = \max\left\{\lceil L \rceil + 1, \left\lceil \frac{1}{\delta}\right\rceil + 1\right\} \in \Bbb{N}.$$ Then $n \ge \left\lceil\frac{1}{\delta}\right\rceil + 1 > \frac{1}{\delta} \implies \delta < \frac{1}{n}$. Further, $n \ge \lceil L \rceil + 1 \ge L + 1 = L + \varepsilon$.

Let $x = 1 + \frac{1}{n}$. Note that $x \in \Bbb{R} \setminus \{1\} = \operatorname{dom} f$. Then, $$|x - 1| = \frac{1}{n} < \delta,$$ and $$|f(x) - L| = \left|\frac{1}{\frac{1}{n} + 1 - 1} - L\right| = |n - L| \ge n - L \ge 1 = \varepsilon.$$ Therefore, $$\forall L \in \Bbb{R}, \exists \varepsilon > 0, \forall \delta > 0, \exists x \in \operatorname{dom} f \text{ such that } |x - 1| < \delta \text{ and } |f(x) - L| \ge \varepsilon.$$