Why can you prove the continuity of $f(x) = \frac{x}{1+x²}$ with $\delta = \min \{ 2, \frac{\epsilon}{2} \} $? Let's take my question as an example. I just don't get it.
What does $\delta = \min \{ 2, \frac{\epsilon}{2} \} $ mean ? (especially 'min{}')

It turns out $\delta=\varepsilon/2$ is plenty in this particular case, so you don't need this sort of argument here.
A better example is to show that $f(x)=x^2$ is continuous at $1$. We do some algebra:
$$|f(1)-f(y)|=|1-y^2|=|1+y||1-y|.$$
Since this has a factor of $|1-y|$, the naive thing to do is to make $\delta$ depend linearly on $\varepsilon$. This has a problem: when $\varepsilon$ is large, $\delta$ will be too large, because $|1+y|$ will grow. For example, taking $\delta=\varepsilon/3$ will not work for $\varepsilon=300$, because for $y=101$ we get $|1-y^2|$ being something like $10000$.
So we need to keep $|1+y|$ moderately small, which we can do only by controlling $|1-y|$. One way is to require $|1-y|<1$ regardless of $\varepsilon$. This works because $|1-y|<1$ is equivalent to $0<y<2$, so we guarantee $1<1+y<3$ and in particular $|1+y|<3$.
This means that if we require $|1-y|<1$ then we have $|1-y^2|<3|1-y|$. We want this to be less than our given $\varepsilon$, so we further require $|1-y|<\varepsilon/3$ to get $|1-y^2|<3 \varepsilon/3 = \varepsilon$ as desired.
To get $|1-y|<1$ and $|1-y|<\varepsilon/3$, we take $\delta=\min \{ 1,\varepsilon/3 \}$.
