Proof that the limit of $\frac{1}{x}$ as $x$ approaches $0$ does not exist

You could argue with the equivalent definition of sequences:

Assume it exists a $L \in \mathbb{R}$ with $\lim_{x \to 0} \frac{1}{x}= L$ then for any sequence $a_n \to 0 $ with $a_n \neq 0$ it follows $\frac{1}{a_n} \to L$.

Now take $a_n = \frac{1}{n}$, then $\frac{1}{a_n} = n \not \to L$ for any $L\in \mathbb{R}$, since the sequence is unbounded (any convergent sequence is bounded).

First assume that $\lim_{x\to 0} {1\over x} = L$. This means that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x$ if $0 < \lvert x\rvert < \delta$ then $\lvert{1\over x} - L \rvert <\epsilon.$

So far, so good.

Now let $\epsilon=1$ and choose $x = \min(1,\delta)$.

Fine, although for which $\delta$? One way to write this would be to say, "$\ldots$ for every $\epsilon > 0$ there is a $\delta_\epsilon > 0$ $\ldots$", so that you can then choose $x = \min(1,\delta_1)$.

Therefore there is a $\delta>0$ such that if $0 < \lvert x\rvert < \delta$ then $\lvert{1\over x} - L \rvert <\epsilon$.

This is confusing: it's not clear why you write "therefore", because we already know this from assuming the limit exists (see the first line, above). Furthermore, in the second line, you have set a value for $\epsilon$ and seem to have implicitly chosen a specific $\delta$, but now you're moving back into generality.

It follows that $\lvert{1\over x}\rvert < 1+ \lvert L\rvert$.

How does that follow?

However clearly $\lvert{1\over x}\rvert>1$ and therefore a contradiction has been reached and there is no number $L$ such that $\lim_{x\to 0} {1\over x} = L$.

Where is the contradiction?