The following map from $\alpha : \mathbb{R} \to \mathbb{R}^2$ is not an embedding
$$ \alpha(t) = \left\{ \begin{array}{ll} (0,-(t+2)) & t \in (-3,-1) \\ \text{regular curve} & t \in (-1,-1/\pi) \\ (-t,-\sin (1/t) & t \in (-1/\pi,0) \end{array} \right. $$
Is not an embedding. And the reason provided is:
Indeed a neighbohood of a point $p$ in the vertical part of the curve consist of an infinite number of connected components in the topology induced by $\mathbb{R}^2$. On the other hand, a neighborhood of such point in the topology "induced" from $\alpha$ (that is the topology of the real line) is an open interval, hence connected.
The curve is:
How do we get the infinite number of connected components? (are these for example straight lines/open segments passing through $p$?). Also I don't get how exactly is proven that $\alpha$ is not an homeomorphism using the connectedness, it is my understanding we would need to prove that a counter-image of a connected set is not a connected set, am I wrong? what exactly is the author proving?
The connected components are not straight lines passing through $p$. If you take a small open ball centered at $p$, then $\alpha\bigl((-3,0)\bigr)$ goes in and goes out of that ball infinitely many times; just recall that $\sin\left(\frac1t\right)=-1$ when $t=\frac1{-\pi/2+2k\pi}$, for some $k\in\mathbb N$.
And so, if $\varepsilon>0$, $\alpha\bigl((-1-\varepsilon,-1+\varepsilon)\bigr)$ is not homeomorphic to $(-1-\varepsilon,-1+\varepsilon)$. But it would be homeomorphic, if $\alpha$ was an embedding.