This is yet another question about the proof on page 340 of Tristan Needham's Visual Complex Analysis (this is the only thing I've read in the entire book as it's the only proof of the crossing rule for winding numbers I could find online). For reference, refer to this image: 
This proof is only intuitional imo. Sure, it's not very hard to write it rigurously if you have the right setting, but this is exactly my problem.
So, suppose we want to prove this statement of the theorem:
Having $\gamma : [0, 1] \to \mathbb{C}$, a continuous closed curve and $z_1, z_2 \in \mathbb{C}\setminus\gamma^*$ two points laying in adjacent connected components of $\mathbb{C}\setminus\gamma^*$, that is, there is a path $\delta : [0, 1] \to \mathbb{C}$ with $\delta(0) = z_1$ and $\delta(1) = z_2$ with the property that $\gamma$ intersects it exactly once, ie. $\exists!(s,t) \in [0, 1]^2:\ \gamma(t) = \delta(s)$,
Then if the orientation of $\gamma$ is such that $z_2$ is on the left of the curve wrt motion down the curve, $I(\gamma, z_1) = I(\gamma, z_2) – 1$. Otherwise, we apply the theorem with $z_1$ and $z_2$ switched and we get that $I(\gamma, z_1) = I(\gamma, z_2) + 1$.
[Notation: $\gamma^* = \gamma([0, 1])$ and $I(\gamma, z)$ is the winding number of $\gamma$ around $z$]
Now, in order for us to be able to use the proof in the book, we need the following setup: the curve $\gamma$ looks like a straight line near the two points $z_1$ and $z_2$. I know that we can "refine" the curve by an homotopy and preserve the winding numbers around the two points, but how can we make sure that they remain in adjacent connected components after this transformation of the curve? In other words, I want to find an explicit way of changing the curve such that after the transformation we can still explicitly find a path linking the two points that $\gamma$ passes through exactly once. Then we can bring the two points as close to the curve as we need along this new path in order to make sure that there is enough space for a circular loop around $z_2$ in its connected component (by using the fact that connected components of complements of paths are open).
Another thing we could do is first transform the curve and then find new points, but then we have the same problem: how do we prove that the winding numbers around our new points (say $z_1', z_2'$) are the same as the ones around the old ones, ie. $I(\gamma, z_1') = I(\gamma, z_1)$ and $I(\gamma, z_2') = I(\gamma, z_2)$.
I've been scratching my head at this for days and I couldn't find a way to make it work. Any help would be appreciated.
Best Answer
I think I finally made it work. It does use topology theory and I admit that I should have added the tag [topology] as well because my question falls a bit outside the area of interest of complex anaysis. (However, I don't have proper knowledge of topological spaces as per se, but topological results I will be using are valid and I have studied in the context of metric spaces, which are just a special case of topological spaces).
First, let $z_1, z_2$ points situated in different connected components of $\mathbb{C}\setminus\gamma^*$ such that there exists a path $\delta$ connecting then and $\exists!(t_0,s_0) \in [0,1)^2:\ \gamma(t_0) = \delta(s_0)=: z_0$. We can assume wlog that $t_0 \ne 0$. Otherwise, we reparametrize the curve such that it starts at some other point $\gamma(t_1)$, $t_1 \in (0, 1)$. Note that such a reparametrization (using the term a bit loosely) still preserves winding numbers.
$z_1 \ne z_0 \ne z_2 \implies |z_1 - z_0| \ne 0$ and $|z_2 - z_0| \ne 0$. Let $0 < R < \min(|z_1 - z_0|, |z_2-z_0|)$. $\gamma$ is continous so $\exists(0 < \eta < \min(t_0, 1-t_0)):\ |t-t_0| < \eta \rightarrow |\gamma(t)-z_0| < R$. Let $a, b$ such that $0 < t_0-\eta < a < t_0 < b < t_0+\eta < 1$. We have that $z_1, z_2 \notin \gamma([a, b]) \subset B(z_0, R)$.
$\gamma([0, a])$ and $\gamma([b, 1])$ are the direct images of compact intervals through a continuous function, hence compact sets in $\mathbb{C}$. From the Heine-Borel Theorem we then have that they are closed sets in $\mathbb{C}$. Their union, which we'll denote by $\gamma^\times$ is therefore closed. Then, $\gamma^\times$ contains all of its accumulation points. Whence, $\gamma(t_0)$ is not an accumulation point of $\gamma^\times$ (remember that we defined $t_0$ such that $\gamma$ crosses it exactly once). This means that we can find a disk of radius $r > 0$ around $z_0$ such that $\overline{B(z_0, r)}\ \cap\ \gamma^\times=\varnothing$.
Next up, we bring the two points $z_1, z_2$ closer to $z_0$ along $\delta$. Since $\mathbb{C}\setminus\mathscr{C}(z_0, r)$ is disconnected, it is also path-disconnected, so $\delta$ must intersect it in order to move from $z_1$ or $z_2 \notin B(z_0, r)$ to $z_0 \in B(z_0, r)$. Let $s_1 = \min\{s \in (0, s_0)\ |\ |\delta(s) - z_0| = r\}$ (we know that this $\min$ exists because the set is the inverse image of a compact set through a continuous function, hence compact). Similarly, let $s_2 = \max\{ s \in (s_0, 1)\ |\ |\delta(s)-z_0|=r \}$. Denote $z_1' = \delta(s_1),\ z_2' = \delta(s_2)$. Then, $z_1', z_2' \in \mathscr C(z_0, r)$. Since $\delta|[0, s_1],\ \delta|[s_2, 1]$ link $z_1, z_1'$ and $z_2, z_2'$ without intersecting $\gamma$, they are respectively in the same connected components, so $I(\gamma, z_i') = I(\gamma, z_i),\ i\in\{1,2\}$.
Let $\delta_1$ be a parametrization of the straight line from $z_1'$ to $z_0$ and similarly $\delta_2$ from $z_0$ to $z_2'$. A disk is convex, so $\delta_1^*, \delta_2^* \in \overline{B(z_0, r)}$.
Define $L$ to be a segment centered at $z_0$ of the bisector line of the angle $\measuredangle z_1z_0z_2$ contained in $B(z_0, r)$. The two segments $\delta_1$ and $\delta_2$ separate $B(z_0, r)$ into 2 sectors. For the two arcs between $z_1'$ and $z_2'$ on $\mathscr C(z_0, r)$, $\gamma$ must intersect both of them because otherwise there would be a path on the circle connecting the two points without intersecting $\gamma$, which would mean that the points are in the same connected component of $\mathbb{C}\setminus{\gamma^*}$, which is false. Moreover, the two arcs without the points $z_1, z_2$ are closed sets in $\mathbb{C}\setminus\{z_1, z_2\}$ and the image of the continuous function $\gamma$ does not contain these two points, so the preimages of the arcs are closed sets in $[0, 1]$, thus bounded, thus from H-B, compact, hence, they admit a minimum and a maximum. For each of the two arcs, there's a first and a last time $\gamma$ crosses them, corresponding to entering and leaving the ball $B(z_0, r)$. I am convinced that $\gamma$ enters the ball through one sector and leaves through the other one, but I'm not quite able to prove this for now. So let $t_{\rm in}, t_{\rm out}$, $0 < t_{\rm in} < t_0 < t_{\rm out}< 1$ be the least and the greatest of the points where $\gamma$ intersects $\mathscr C(z_0, r)$.
Define $z_{\rm in}, z_{\rm out}$ as respectively the images of $t_{\rm in}, t_{\rm out}$ through $\gamma$. Denote the endpoints of $L$ as $l_{\rm in}$ and $l_{\rm out}$ according to which sector they are in ($t_{\rm in/out}$ in the same sector as $l_{\rm in/out}$).
For two points $x,y$ on the complex plane, we denote by $[x, y]$ a parametrisation of the line segment from $x$ to $y$ (while keeping the usual meaning for $x, y \in \mathbb{R}$). Also, we use $\cup$ for path concatenation. Now consider the closed curves:
$G_{\rm in} = \gamma\big|[a, t_{\rm in}] \cup [z_{\rm in}, l_{\rm in}] \cup [l_{\rm in}, z_0]$
$G_{\rm out} = [z_0, l_{\rm out}] \cup [l_{\rm out}, z_{\rm out}] \cup \gamma\big|[t_{\rm out}, b]$
$\gamma([a, t_0]), G_{\rm in}, \gamma([t_0, b]), G_{\rm out} \subset B(z_0, R) \not\ni z_1, z_2$ and $B(z_0, R)$ is simply-connected $\implies$ $\gamma([a, t_0]) \sim G_{\rm in}$ and $\gamma([t_0, b]) \sim G_{\rm out}$ are homotopic in $\mathbb C\setminus\{z_1, z_2\}$. We can replace the curve $\gamma$ with the curve $\overline\gamma := \gamma\big|[0, a] \cup G_{\rm in} \cup G_{\rm out} \cup \gamma\big|[b, 1]$. $\gamma \sim \overline\gamma$ in $\mathbb C\setminus\{z_1, z_2\}$, therefore $I(\overline\gamma, z_{1/2}) = I(\gamma, z_{1/2})$. Moreover, $\overline{\gamma}^* \setminus \overline{B(z_0, r)} \subseteq \gamma^* \setminus \overline{B(z_0, r)}$, so $\delta([0, s_0]), \delta([s_1, 1]) \notin \overline{\gamma}^*$. This means that $z_1', z_1$ and $z_2', z_2$ are still respectively in the same connected components of $\mathbb C\setminus \overline{\gamma}^*$. Then $I(\overline\gamma, z_{1/2}') = I(\overline\gamma, z_{1/2}) = I(\gamma, z_{1/2})$.
From this point onward we can use the method of proof presented in the refferenced book.
Via a geometric argument we see that there is a ball $B(z_0, r') \subset B(z_0, r)$ that is disjoint from $[l_{\rm in/out}, z_{\rm in/out}]$. Altenatively, $B(z_0, r)$ minus the two segments is an open set and $z_0$ is inside it. The boundary of this ball, $\mathscr C(z_0, r')$ is cut in half by $L$ and $z_1', z_2'$ are on either side of it. If we take the half that is intersected by $[z_0, z_2']$, we can replace the portion of $L$ that is inside $B(z_0, r')$ by it via an homotopy because both curves are inside $B(z_0, r)$ which is simply-connected and doesn't contain $z_1', z_2'$. Call this new curve $\widetilde{\gamma}$. This has the same winding numbers around $z_0, z_2'$ as $\overline\gamma$ around $z_1', z_2'$.
Choose a point $z_3$ on $[z_0, z_2']^* \cap B(z_0, r')$ close enough to $z_0$ that there is a circle around it, $\mathscr C(z_3, \rho)$, that is tangent to $L$ and is contained inside this half of $B(z_0, r')$. We have $I(\tilde\gamma, z_3) = I(\tilde\gamma, z_0) = I(\overline\gamma, z_1')$ and $I(\tilde\gamma, z_2') = I(\overline\gamma, z_2') = I(\overline\gamma, z_3)$. Construct the curve $\mathring\gamma$ by taking $\overline\gamma$ up to $z_0$ followed by the boundary of $B(z_3, \rho)$ traversed in opposite direction to $L$ (the opposite sense to the one of traversing $\mathscr C(z_0, r)$ going through $z_{\rm in}, z_1'$ and $z_{\rm out}$ in this order), then the rest of $\overline\gamma$. It is not very hard to see that $\tilde\gamma \sim \mathring\gamma$ in $\mathbb C \setminus \{z_3, z_2'\}$ (We can build the homotopy by considering segments from the intersection point of $\mathscr C(z_3, \rho)$ with $L$ to points on the semi-circle around it from $\tilde\gamma$ and dragging those points to the intersections of the segments with $\mathscr C(z_3, \rho)$ along the segments, with the same speed. At the same time, we extend either side of $L \setminus \overline{B(z_0, r')}$ up to where we drag the points $L \cap \mathscr C(z_0, r')$.)
This all means that $I(\gamma, z_1) = I(\mathring\gamma, z_3)$ and $I(\gamma, z_2) = I(\overline\gamma, z_3)$. Via a lemma that is easy to prove, as $\mathring\gamma$ is the concatenation of $\overline\gamma$ with $C = \partial B(z_3, \rho)$ under a change of starting point to $z_0$, $I(\mathring\gamma, z_3) = I(\overline\gamma, z_3) + I(C, z_3) \implies I(\gamma, z_1) = I(\gamma, z_2) + I(C, z3)$. Depending on the orientation of $C$, $I(C, z_3) = \pm 1$. If $\gamma$ is oriented such that $z_1$ is on its left, then $z_1'$ is on the left of $L$ (I'm not too sure whether "$z_1$/the connecetd component of $z_1$ is on the left side of $\gamma$" makes much sense generally, but in cases where it doesn't, I would argue that this works as a definition) which means that $z_{\rm in}, z_1'$ and $z_{\rm out}$ are traversed in this order clockwise, so $C$ is traversed counter-clockwise. This being said, we can finally conclude that $I(\gamma, z_1) = I(\gamma, z_2) - 1$.
Sidenote: I've gone through the last part of the proof (the one adapted from the proof in Tristan Needham's book) in a faster pace because, even though it is fundamental to the proof as a whole, my question was about the setup I got before that part, and also because I've been writing on this proof for more than a week now and I really want to call it quits and not waste any more time on it. If anyone could help me complete the missing detail from the first part, I'd be thankful.