Your example is not totally clear but I think I see what you missed. In Hatchers statement you notice he only writes $\pi_1(X)$ and not $\pi_1(X,x_0)$. This due to the fact that his hypothesises imply that $X$ is path connected.
This last thing implies that the fundamental group at each point is the same (up to isomorphism) : Suppose that $X$ is path connected one has a clear isomorphism between $\pi_1(X,x_0)$ and $\pi_1(X,y_0)$ for all $x_0,y_0 \in X$. Just choose a path $\gamma : [0;1]\longrightarrow X$ such that $\gamma(0)=x_0$ and $ \gamma(1)=y_0$. Then you have the isomorphism $\phi_\gamma : \pi_1(X,x_0)\longrightarrow \pi_1(X,y_0)$ given by $\phi_\gamma([l]) = [\gamma * l * \gamma^{-1}]$.
So finally, even if you have to take care that such a point $x_0$ and a covering of $X$ path-connected sets of the form $(A_\alpha)_\alpha$ as in the theorem exist in order to apply it, since each set involved in the statement is path connected and contains $x_0$ you can talk about the fundamental group of such a set (which is the fundamental group of this set at $x_0$ a priori but doesn't depend on the choice of the point by path-connectedness, so it depends only on the set considered).
I hope this answers your question !
Reducing the nlab-theorem to tom Dieck's theorem breaks down when one tries to show that the interiors of $\tilde X_0, \tilde X_1$ cover $\tilde X$. At least there is no simple proof - but nevertheless it could be true. Anyway, we do not need it. In fact, tom Dieck's theorem relies on two ingredients:
Theorem (2.6.1) which states a pushout property for fundamental groupoids under the assumption that $X_0$ and $X_1$ are subspaces of $X$ such that the
interiors cover $X$.
The existence of a retraction functor $r : \Pi(Z) \to \Pi(Z,z)$ which tom Dieck only defines for path connected $Z$. This works as follows: For each object $x$ of $\Pi(Z)$ (i.e. each point $x \in Z$) we define $r(x) = z$. For the morphisms we proceed as follows: We choose any morphism $u_x : x \to z$ if $x \ne z$ and take $u_z = id_z$= path homotopy class of the constant path at $z$. Given a morphism $\alpha : x \to y$ in $\Pi(Z)$, we define $r(\alpha) = u_y \alpha u_x^{-1}$.
We shall see that 2. can be generalized so that we can prove
Theorem (Seifert - van Kampen). Let $X$ be a topological space and $X_0,X_1\subset X$ be subsets whose interiors cover $X$ such that $X_{01}=X_0\cap X_1$ is path connected. Then for any choice of base point $\ast\in X_{01}$
\begin{matrix}\pi_1(X_{01},\ast) & \to & \pi_1(X_0,\ast)\\
\downarrow&&\downarrow \\
\pi_1(X_1,\ast) & \to & \pi_1(X,\ast)
\end{matrix}
is a pushout in the category of groups.
Proof. As Tyrone suggested in his comment, for a pointed space $(Z,z)$ let us denote by $\tilde Z$ the path-component of $Z$ containing the basepoint $z$. From Jackson's and your answers we know that for $X_{01}$ path connected and $* \in X_{01}$ we have $\tilde X_{01} := \tilde X_0 \cap \tilde X_1 = X_{01}$ and $\tilde X = \tilde X_0 \cup \tilde X_1$. Note that $X_{01}$ path connected is essential for both equations.
We apply tom Dieck's retraction contruction to $Z = \tilde X$ and $z = * \in X_{01} = \tilde X_{01}$ by first chosing $u_x$ in $\Pi(X_{01})$ for all $x \in X_{01}$ (where of course $u_* = id_*$), then $u_x$ in $\Pi(\tilde X_0)$ for all $x \in \tilde X_0 \setminus X_{01}$ and finally $u_x$ in $\Pi(\tilde X_1)$ for all $x \in \tilde X_1 \setminus X_{01}$. Since $\Pi(X_{01}),\Pi(\tilde X_0), \Pi(\tilde X_1)$ are subcategories of $\Pi(\tilde X)$, this gives us a choice of $u_x$ in $\Pi(\tilde X)$ for all $x \in \tilde X$ providing a retraction $\tilde r : \Pi(\tilde X) \to \Pi(\tilde X,*) = \Pi(X,*)$. We extend it to a retraction $r : \Pi(X) \to \Pi(X,*)$ as follows: Given a morphisms $\alpha : x \to y$ in $\Pi(x)$, then $x,y$ belong to same path component $P$ of $X$. If $P = \tilde X$, we define $r(\alpha) = \tilde r(\alpha)$. If $P \ne \tilde X$, we define $r(\alpha) =id_*$. Consider the restriction $r_{01}: \Pi(X_{01}) \to \Pi(X,*)$. The category $\Pi(X_{01})$ is a subcategory of $\Pi(\tilde X,*)$ and by construction $r_{01}(\Pi(X_{01})) = \tilde r(\Pi(X_{01})) \subset \Pi(X_{01},*)$, i.e. we may regard $r_{01}$ as a map $r_{01} : \Pi(X_{01}) \to \Pi(X_{01},*)$. Next consider the restriction $r_0 : \Pi(X_0) \to \Pi(X,*)$. For the subcategory $\Pi(\tilde X_0) \subset \Pi(X_0)$ we have by construction $r_0(\Pi(\tilde X_0)) = \tilde r(\Pi(\tilde X_0)) \subset \Pi(X_0,*)$. Let $\tilde P_0$ be a path component of $X_0$ different from $\tilde X_0$, i.e. $\tilde P_0 \cap \tilde X_0 = \emptyset$. Then $\tilde P_0 \cap \tilde X = \tilde P_0 \cap \tilde X_1 = \tilde P_0 \cap X_0 \cap \tilde X_1 \subset \tilde P_0 \cap X_0 \cap X_1 = \tilde P_0 \cap \tilde X_0 \cap \tilde X_1 \subset \tilde P_0 \cap \tilde X_0 = \emptyset$. Thus $\tilde P_0 \cap \tilde X = \emptyset$ and therefore by construction $r_0(\Pi(\tilde P_0)) = r(\Pi(\tilde P_0) = \{id_*\} \subset \Pi(X_0,*)$. We conclude $r_0(\Pi(X_0)) \subset \Pi(X_0,*)$, i.e. we may regard $r_0$ as a map $r_0 : \Pi(X_0) \to \Pi(X_0,*)$. Similarly $r$ restricts to $r_1 : \Pi(X_1) \to \Pi(X_1,*)$. Therefore we get a commutative diagram
\begin{matrix}\Pi(X_0) & \hookleftarrow & \Pi_1(X_{01}) & \hookrightarrow & \Pi(X_1)\\
\downarrow r_0 && \downarrow r_{01} &&\downarrow r_1 \\
\Pi(X_0,*) & \hookleftarrow & \Pi(X_{01},*) & \hookrightarrow & \Pi(X_1,*)
\end{matrix}
Now the same argument as in the proof of tom Dieck's Theorem (2.6.2) applies.
Best Answer
Well...the first thing you're missing is that your $n$ might not be finite unless you have some nice assumptions. For instance, if $A$ is the closed upper half-plane, and $B$ is the closed lower half-plane, then the graph of $$ f(x) = \begin{cases} x^2 \sin \frac{1}{x} & x \ne 0\\ 0 & x = 0 \end{cases} $$ has infinitely many arcs in $A$ and in $B$. "But," I hear you cry, "we have that $A$ and $B$ are open!" Sure you do...and is that enough to prove finiteness of $n$? Well...yes, kind of, but there's work to do...and there's a half-page of your life gone.
"But after that," you say, "The only subtle thing is getting rid of the equivalent things ... that "amalgamation" stuff, right?" Well... yes and no. Showing that curves that are equivalent modulo the relevant stuff in the amalgamated product are in fact homotopic is subtle. But you also have to prove that those are the ONLY ones that are, and that's subtle too. And there's another page or two.
You might want to remember that once you have the van Kampen theorem, you can prove the Jordan Curve Theorem, for which a correct (and understood-to-be-correct) proof took a very long time. So...it's not likely to be an easy proof.