$\varepsilon$-$\delta$ proofs seem to be the most confusing concept of first-year Calculus (or pre-Calculus, depending on how it's taught).
The $\varepsilon$-$\delta$ definition of a limit is as follows. $\lim_{x\rightarrow a} f(x) = l$ means for all $\varepsilon > 0$, there is a $\delta > 0$ such that if $|x - a| < \delta$ then $|f(x) - l| < \varepsilon$. All your answers are contained within this definition so let's break it down and see how all the proofs you reference satisfy the definition.
For the limit to be $l$, we require that every $\epsilon > 0$ has a corresponding $\delta > 0$. A standard way to prove that something holds for all values of a variable in a certain domain is to choose an arbitrary (unspecified) value for that variable in its domain and make a proof that works for that arbitrary choice. Spivak's proof does exactly that: he says "for any $\varepsilon > 0$" and then proceeds to show that there is a $\delta$ corresponding to this specific but arbitrary $\varepsilon$. The unstated conclusion is that there is a $\delta$ for every $\varepsilon$.
Note that the proof of a limit requires that there is a $\delta$ corresponding to every positive "epsilon", but there is no requirement to denote the "epsilon" value as the symbol $\varepsilon$. Instead, we may use an expression such as $c\varepsilon$ for constant $c>0$ so long as this expression can take on all values in the domain of the "epsilon" (namely, all positive numbers). Clearly we may write any positive number as $c\varepsilon$ for some choice of $\varepsilon$. So if we prove that a $\delta$ exists for all $c\varepsilon$ in place of $\varepsilon$, the proof holds for every value that $c\varepsilon$ can take, which means it holds for every positive number, so we are good.
However, if the expression we choose can't take on all possible "epsilon" values, such as $1+\varepsilon$ for $\varepsilon > 0$, then if we prove that a $\delta$ exists for all $1+\varepsilon$ then we haven't proven that a $\delta$ exists for all values for which the definition of limit requires it exist. Such a proof would be invalid.
Finally, you posted a link to another question where a complex expression $\min\left(1,\frac{\epsilon}{2(|m|+1)}\right)$ stands in for the "epsilon". Note that by letting $\epsilon$ vary, this expression can take on all positive values up to but not exceeding 1. Therefore it doesn't take on all values that the definition of limit requires. However, if a $\delta$ works for some "epsilon" then it automatically works for any larger "epsilon" - you can immediately verify this statement in the definition of a limit. So if our expression can take on all positive numbers up to 1, that's actually good enough. (In the most generality, we can prove that there is a $\delta$ corresponding to all "epsilon" in a set of positive numbers containing a sequence going to 0.)
Best Answer
Let $\varepsilon>0$ be fixed. Since $f(x)\to\pi$ as $x\to 0$, there is some $\delta'>0$ such that
$$\left|f(x)-\pi\right|<\varepsilon$$
whenever $\left|x\right|<\delta'$. Similarly, since $\sin(x)\to0$ as $x\to\pi$, there is some $\delta>0$ such that
$$\left|\sin(x)\right|<\delta'$$
whenever $\left|x-\pi\right|<\delta$. Combining these we notice that we have the implications
$$\left|x-\pi\right|<\delta \implies \left|\sin(x)\right|<\delta' \implies \left|f(\sin(x))-\pi\right|<\varepsilon.$$
Thus
$$\left|f(\sin(x))-\pi\right|<\varepsilon$$
whenever $\left|x-\pi\right|<\delta$, which proves the assertion.