Mike Reid's Undergraduate Algebraic Geometry gives a discussion of the topology (as a subspace of $\Bbb{P}^2_{\Bbb{C}}$) of a cubic curve $C : y^2 = q(x) \cup \{\infty\}$, along the same lines as the answer by user8268 to this MSE question how is a cubic curve topologically equivalent to a torus? Both discussions describe the geometry nicely but don't give details of the proof of the main claim.
Reid takes $q(x) = x(x-1)(x-\lambda)$. He considers the function $\pi : C \to \Bbb{P}^1_{\Bbb{C}}$ given in affine coordinates by $(x, y) \mapsto x$ and $\infty \mapsto \infty$. $\pi$ is a double covering except over $\{0, 1, \lambda, \infty\}$. He then considers disjoint paths $01$ and $\lambda\infty$ with the indicated endpoints. Reid claims that if the parts of $C$ lying over these paths is removed, $C$ falls into two pieces. I.e., if we call the two paths $M$ and $N$, he is claiming that $E = C \setminus \pi^{-1}[M \cup N]$ has two connected components.
After quite a long struggle, I believe I can now prove this claim using the theory of covering projections. Here is a sketch of my proof. It is fairly easy to see that $\pi\mid_{E}$ is a (double) covering projection. Since the base $B = \Bbb{P}^1_{\Bbb{C}} \setminus (M \cup N)$ of this double cover is homeomorphic to $S^1 \times (0, 1)$, there are two possibilities: either $E$ is homeomorphic to $B$ and the double cover is induced by the double cover $z \mapsto z^2$ of $S^1$ or $E$ is homeomorphic to the disjoint union $B \sqcup B$ and the double cover is the trivial one that restricts to the identity on each component. What Reid is claiming is that the latter holds, so that there is a continuous map sending $x \in B$ to a $y$ such that $(x, y), (x, -y) \in C$, i.e., we have a continuous choice for $\sqrt{q(x)}$. To see that this holds, take $\alpha : S^1 \to B$ to be a generator of the fundamental group $\pi_1(B)$. We must show that $q \circ \alpha$ lifts to the double cover $z \mapsto z^2$ of $\Bbb{C} \setminus \{0\}$ (so that the lifting provides a value for $\sqrt{q(\alpha(s))}$ as a continuous function of $s \in S^1$). Now parametrise the paths $M$ and $N$ by continuous maps $\mu, \nu: [0, 1] \to \Bbb{P}^1_{\Bbb{C}}$ satisfying:
$$\mu(0) = 1 \quad \mu(1) = 0 \quad \nu(0) = \lambda \quad\nu(1) = \infty$$
and, for $t \in [0, 1]$, put:
$$ q_t(x) = x (x - \mu(t)) \left(\frac{\lambda x}{\nu(t)} - \lambda\right) $$
Then $q_t(x)$ has no roots on $B$ and hence we have a homotopy between $q = q_0$ and $q_1$ as maps $B \to \Bbb{C} \setminus \{0\}$. But $q_1(z) = -\lambda z^2$. So $q \circ \alpha$ has degree $2$ and does indeed lift to the double cover of $\Bbb{C} \setminus \{0\}$.
I am interested to know whether my proof is the usual one and what alternatives there are. I would appreciate any relevant pointers to the literature. I am guessing that the above is a special case of some general result about the degree of $q \circ \alpha : S^1 \to \Bbb{C} \setminus \{0\}$, where $q$ is polynomial of degree $n$ and $\alpha$ is a curve that winds once around $m$ of the $n$ roots of $q$.