Unclear about the epsilon-delta definition of continuous mapping

In short, I don't see how the epsilon-delta definition excludes non-injective mappings. For example, I can imagine a modified sine where a single point is excluded, but for which we can still find the finite neighbourhood $$\delta$$.

Your drawing seems way wrong. You are given $$\epsilon$$, and you then have to come up with a $$\delta$$ so that the entire $$\delta$$-ball around $$x_0$$ is mapped into the $$\epsilon$$-ball around $$f(x_0)$$. If $$\epsilon$$ is sufficiently small, you won't be able to do that with the graph you show since there will be points arbitrarily close to $$x_0$$ which are mapped "far" (more than $$\epsilon$$ away) from $$f(x_0)$$.

In your graph, a tiny ball around $$x_0$$ will be mapped to a tiny ball around the y-value of the "missing" point on the graph, except for the single value $$f(x_0)$$. Except for the single point, the image of the ball will be far from $$f(x_0)$$.

There are some things you are confusing. First and foremost, injectivity has nothing to do with continuity.

$$f : \mathbb{R} \to \mathbb{R}, x\mapsto x^2$$ is continous but neither injective nor surjective.

And then i suppose there's some confusion about the $$\varepsilon, \delta$$-criteria.

I'm not sure what you meant by excluding a single point of your function. However, what you certainly can do, roughly speaking, is tearing the graph of a function apart by some piecewise definition such as

$$f : \mathbb{R} \to \mathbb{R}, x\mapsto \begin{cases} \sin(x) & \ x < 1 \\ 0 & \ x = 1 \\ \sin(x) & \ 1 < x \end{cases}$$

And indeed you can always give me an $$\varepsilon$$ for which there is a $$\delta$$ such that the $$\varepsilon,\delta$$ criteria holds.

But the key is that there is at least one $$\varepsilon'$$ around $$f(1)$$ for which it fails and thus the given function is not continous.

Long story short: It's not about finding one $$\varepsilon > 0$$ for which you can find a $$\delta$$ such that

$$\vert x-x'\vert < \delta \Rightarrow \vert f(x) - f(x') \vert < \varepsilon$$

You must be able to find such $$\delta$$ for every $$\varepsilon > 0$$ for which the implication of the $$\varepsilon,\delta$$ criteria holds.