Let's show that
$$
||x|-|y|| \leq |x-y|
$$
for any $x,y \in \mathbb{R}$. Indeed, by the triangle inequality, we have
\begin{align}
|x| &= |x - y + y| \\
&\leq |x-y| + |y|,
\end{align}
which implies $|x| - |y| \leq |x-y|$. Exchanging $x$ and $y$ leads to $|y| - |x| \leq |x-y|$. Hence, the inequality is established. Can you simplify your proof from here?
i.e. I wanted to write a proof that:
$$\forall L \in \mathbb R \Big [\exists \epsilon \gt 0 \text{ s.t. } \forall N \in \mathbb R \ \exists x \in \mathbb R \text{ s.t. }\big (x \gt N \land \lvert \sin(x) - L \rvert \geq \epsilon \big) \Big]$$
We can do this using $\varepsilon = \frac{1}{2}$.
Case 1: $L > 0$
Here for any $N$ we can always find $x > N$ of the form $x = \frac{3\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
For such $x$ we have
$$\sin x = -1.$$
Since $L > 0$ we have
\begin{align}
|\sin x - L|&= |-1 - L| \\
& = 1 + L \\
& > \frac{1}{2}
\end{align}
Case 2 $L < 0$
This case is nearly identical to the case for $L > 0$.
For any $N$ we can always find $x > N$ of the form $x = \frac{\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
For such $x$ we have
$$\sin x = 1.$$
Since $L < 0$ we have
\begin{align}
|\sin x - L|&= |1 - L| \\
& = 1 + |L| \\
& > \frac{1}{2}
\end{align}
Case 3 $L = 0$
This case can be dealt with immediately by re-using either of the $x$ from the previous two cases:
If $x = \frac{\pi}{2} + 2k\pi$, then
\begin{align}
|\sin x - L|&= |1 - 0| \\
& = 1 \\
& > \frac{1}{2}
\end{align}
Of course, this all depends on knowing the behavior of $\sin$: that $\sin$ wavers between $1$ and $-1$ "forever" and so on. Given this behavior, one would expect intuitively that the limit doesn't exist, i.e. $\sin x$ wouldn't "settle" upon some single value for large $x$.
Later in the book Spivak defines the trigonometric functions formally in a way that conforms with the geometric/unit circle versions with which you are probably already familiar. Continuity plays a big part in these later definitions.
Here though, the main thing is just the periodic nature of $\sin$.
Finally, you may be somewhat leery of statements like:
...for any $N$ we can always find $x > N$ of the form $x = \frac{3\pi}{2} + 2k\pi$ where $k \in \mathbb{N}$.
This has at its heart the idea that given any real number $N$ we can find a natural number $n$ with $n > N$. This fact is employed in a few places early on in the text, but isn't formally proven until Chapter 8 (3rd Ed.) "Least Upper Bounds".
Spivak cops to this "cheating" at the end of that chapter.
(As an aside, an earlier problem in Chapter 5, 5-12 gives a very useful result:
If $f(x) \leq g(x)$ for all $x$ (or even just for all $x$ in a neighborhood of $a$) then $\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)$, provided these limits exist.
This can be shown to hold if we replace $a$ with $\infty$.
This, combined with your lemma that $|\sin x| \leq 1$ can be used to rule out your Case 1, i.e. since $-1 \leq \sin x \leq 1$ we must also have $-1 \leq \lim_{x \to \infty}\sin x \leq 1$, if that limit exists.)
Best Answer
The error lies in the sentence
As I commented, the choice of $\delta$ cannot depend on $x$ because there is no variable $x$ in scope at this point.
Let's review the strategy for crafting an $\epsilon$-$\delta$ continuity proof. Given the function $$ f(x) = x^2, \quad x \in \mathbb{R}, $$ our goal is to show that
which is by definition equivalent to
Fix $a \in \mathbb{R}$. By the definition of continuity, we need to show that
Fix $\epsilon > 0$. To show that there exists a number $\delta > 0$ such that the aforementioned condition holds, we pick a value for $\delta$ based on quantities that are in scope (namely $a$ and $\epsilon$) and show that the desired condition holds for that value of $\delta$.
To do so, we observe that the desired relation $\lvert x^2 - a^2 \rvert < \epsilon$ is equivalent to $\lvert x - a \rvert \lvert x + a \rvert < \epsilon$, so we would like to bound both factors $\lvert x - a \rvert$ and $\lvert x + a \rvert$ from above. Automatically, the first factor $\lvert x - a \rvert$ is bound by whatever value of $\delta$ we choose. There is some freedom in how we choose to bound the second factor $\lvert x + a \rvert$; one possibility is to take advantage of the triangle inequality to get $$ \lvert x + a \rvert = \lvert (x - a) + 2a \rvert \le \lvert x - a \rvert + \lvert 2a \rvert < \delta + 2 \lvert a \rvert. $$ If we choose $\delta$ such that, for example, $$ \delta \le 1, \tag{1} \label{1} $$ then $$ \lvert x + a \rvert < 1 + 2 \lvert a \rvert, $$ so that in order to guarantee $\lvert x - a \rvert \lvert x + a \rvert < \epsilon$, it suffices to make $$ \lvert x - a \rvert < \frac{\epsilon}{1 + 2 \lvert a \rvert}. \tag{2} \label{2} $$ Therefore, we decide to pick $$ \delta = \min \Bigl\{ 1, \frac{\epsilon}{1 + 2 \lvert a \rvert} \Bigr\} $$ in order to guarantee both \eqref{1} and \eqref{2}.
This is, of course, not the only possible choice of $\delta$. For example, Thomas Andrews's answer mentions an alternative choice $$ \delta = \sqrt{\epsilon + a^2} - \lvert a \rvert. $$