The non-existence of $\lim \limits_{x \to 0} \sin {1 \over x}$

Hint: Either $L=1$ or $L\neq1$. In the former case, when do we have $\sin(x)=-1$ for $x\in\Bbb R^+$? (Think setwise, not specific values). Now invert those numbers, and see that you have found infinitely many places in $(0,1)$ with $|\sin(1/x)-L|>1-\varepsilon$ for any $\varepsilon>0$. In the second case, instead solve for $\sin(x)=1$, and repeat a similar process.

Your approach is almost correct, but the order in which you introduce your variables is confusing and may cause trouble.

You don't just find an $x$ and an $\varepsilon$ which may depend on $L$. Of course $\varepsilon$ may depend on $L$, but $x$ can depend both on $L$ and on $\delta$.

I find it sometimes helps to think of problems like this as an adversarial game, like chess. My opponent will make the first "move", choosing $L$. Then it is my "move": I must choose a value of $\varepsilon$. After that, my opponent's "move" is to choose $\delta$. I must then "move" by choosing an $x$ such that $0 < \left| x \right| < \delta \wedge \left| {\sin {1 \over x} - L} \right| \ge \varepsilon $, and then I win.

The pattern this game follows is that you read off the quantifiers from the outside, working inward (left to right). Every universal $\forall$ is a "move" by my opponent, and every existential $\exists$ is one of my "moves."

The problem is to show a perfect "strategy" that will win any game no matter what "moves" my opponent makes.

Often it is helpful to give a strategy that handles some values of $L$ one way and some values a different way. In another answer you have a strategy that treats $L=1$ as one case and $L\neq 1$ as a different case. My first instinct looking at this problem was to come up with a strategy for the case $L \geq 0$, and then look at the strategy for the case $L < 0$. The two strategies are almost the same, as it turns out.

Edit: Here are more details. What follows is not a detailed proof (I want to leave some of this as an exercise!) but some very strong hints.

The intuition behind this comes from the graph of $\sin\frac1x$, which oscillates between $1$ and $-1$ infinitely often as $x$ approaches $0$ from either side. The condition $0 < |x| < \delta$ allows us to use positive or negative values of $x$, but the positive values will be more than enough for our needs. Note that $\sin u = 1$ for $u=\frac\pi2, \frac\pi2 + 2\pi, \frac\pi2 + 4\pi, \ldots$, while $\sin u = -1$ for $u=\frac{3\pi}2, \frac{3\pi}2 + 2\pi, \ldots$. Let $x = \frac1u$. If $u > \frac1\delta$, then $0 < x < \delta$. So no matter how the opponent chooses $\delta$, as long as $\delta > 0$ there are plenty of ways to choose $x$ so that $\sin\frac1x = 1$ and $0 < x < \delta$, or to choose $x$ so that $\sin\frac1x = -1$ and $0 < x < \delta$.

So on the last "move", we'll always be able to choose between $\sin\frac1x = 1$ and $\sin\frac1x = -1$, whichever we want. If $L \geq 0$ and $\sin\frac1x = -1$, then $\left| \sin\frac1x - L \right| \geq \underline{\phantom{00}}$. If $L < 0$ and $\sin\frac1x = 1$ , then $\left| \sin\frac1x - L \right| > \underline{\phantom{00}}$. Fill in the blanks in those statements, and then it should not be hard to think of an $\varepsilon$ that will work in each case, that is, so that $\left| \sin\frac1x - L \right| > \varepsilon$. As it turns out, $\varepsilon$ does not need to be a function of $L$ for this particular proof.

Having devised a strategy, all that remains is to prove, step by step, that the strategy will "win" for any $L \geq 0$, and also that it will "win" for any $L < 0$.

Many ways and hints you've got are useful. This solution is good too:

Lemma1: The limit of a real function is unique if it exists.
Lemma2: The limit of a real function $f$ at $x=a$ is $L$ iff for any sequence $(a_n)$ which converges to $a$, sequence $f(a_n)$ converges to $L$.

So if $\sin {1 \over x}$ has limit at $x=0$, for two sequences $(a_n)=({1\over n\pi})$ and $(b_n)=({2\over (4n+1)\pi})$ we must have: $$\lim\limits_{n \to \infty} \sin {1 \over a_n}=\ \lim\limits_{n \to \infty} \sin {1 \over b_n}$$ Because both are the limit. Simple calculations show that LHS is $0$ and RHS is $1$, a contradiction.