If you want a result without groups or groupoids, here it is.
Let $\cal U$ be an open cover of a space $X$ with base point $x$ such that each set of $\cal U$ is simply connected and each set of $\cal U$ and intersection of two sets of $\cal U$ is path connected and contains $x$. Then $X$ is simply connected.
Proof. Let $a: I \to X $ be a loop in $X $ at the base point $x $. By the Lebesgue covering lemma, there is a subdivision $ a=a_1 + a_2+ \cdots +a_n $ of $a $ such that each $a_i $ lies in some set $U_i $ of $\cal U$. We consider $a_i$ as a map $I \to X$. Let $b_0, b_{n+1}$ be constant paths at the point $x$. By the assumptions, we can for $1 < i \leq n$ choose a path $b_i: I \to X$ joining $a_i(0)$ to $x$ and lying in $U_{i-1} \cap U_i$. The loop $-b_i + a_i+b_{i+1}$ lies in $U_i$ and so is contractible to a constant rel end points, by assumption of simple connectivity of $U_i$. It follows that $a$ is contractible to a point in $X$.
This argument can be usefully generalised to higher dimensions, in the context of filtered spaces.
Do it by induction on $n$. The case $n = 2$ is exactly Van Kampen's theorem: $X_1$ and $X_2$ are path-connected, and so is their intersection (it's the intersection of two convex sets, hence convex, and it's nonempty, hence path-connected). Thus
$$\pi_1(X_1 \cup X_2) \cong \pi_1(X_1) *_{\pi_1(X_1 \cap X_2)} \pi_1(X_2) = 0.$$
Now for the induction step, suppose that you know the result is true for a given $n \ge 2$ and let $X_1, \dots, X_{n+1}$ be convex sets such that $X_i \cap X_j \cap X_k$ is connected for all $i,j,k$. By the induction hypothesis, $Y = X_1 \cup \dots \cup X_n$ is simply connected. It remains to show that $Y \cap X_{n+1}$ is path-connected, and you can apply Van Kampen's theorem again to conclude. (1)
So let's show $Y \cap X_{n+1}$ is path-connected. Clearly
$$Y \cap X_{n+1} = (X_1 \cap X_{n+1}) \cup \dots \cup (X_n \cap X_{n+1}).$$
Now suppose $x,y$ belong to $Y$, we are looking for a path from $x$ to $y$. Let
$$x \in X_i \cap X_{n+1}, \qquad y \in X_j \cap X_{n+1}.$$ By the hypothesis on the $X_\cdot$, the intersection $X_i \cap X_j \cap X_{n+1}$ is non-empty; choose $z$ inside it. Since $X_i \cap X_{n+1}$ is path-connected (it's the nonempty intersection of two convex sets), there is a path $\gamma$ from $x$ to $z$. Similarly, there is a path $\gamma'$ from $z$ to $y$. Concatenating these two path gives a path $\alpha = \gamma \cdot \gamma'$ from $x$ to $y$. Thus $Y \cap X_{n+1}$ is path-connected and we can conclude (cf. (1)).
Remark. You forgot the assumption (which is written in Hatcher's book) that the sets $X_i$ have to be open. This is crucial to apply van Kampen's theorem, in general it's false.
Best Answer
Here are a couple of bounded examples.
Let $U=\{(x,y):0<x<1, 0<y<x^2\}.$
Let $P=\{(x,y):1<x^2+y^2<2, 0<y\}.$
The proofs are similar, here is a proof for $P$. Let $C=\{(x,y):1=x^2+y^2, 0\le y\}.$ If $V$ is any convex open subset of $P$ then its closure $K$ is a compact convex subset which intersects $C$ in at most one point. If $V_1,...,V_n$ are finitely many convex open subset of $P$, then their closures $K_1,...,K_n$ intersect $C$ in at most finitely many points, and hence there is some $(p,q)\in C\setminus \cup_{i=1}^n K_i.$ The set $K=\cup_{i=1}^n K_i$ is compact, hence the distance from $(p,q)$ to $K$ is positive, hence there are points $(x,y)$ near $(p,q)$ in $P\setminus K$, so such $(x,y)$ are not covered by any of the $V_1,...,V_n.$