$\newcommand{\R}{\mathbb R}$ $\newcommand{\Z}{\mathbb Z}$ $\DeclareMathOperator{\im}{Im}$ View $S^1$ as the boundary of the $2$-disk $D^2=\{(x,y)\in \R^2 : x^2 + y^2 \leq 1\}$. Fix $x_0 \in S^1$ and let $X=(S^1 \times S^1) \cup (D^2 \times \{x_0\})$. Compute $\pi_1 (X)$.
Let $X=A\cup B$ where $A=(S^1 \times S^1)$ and $B= D^2 \times \{x_0\}$. Intuitively, $X$ can be thought of as a torus with a disc glued inside it, so that the intersection of of the $2$-torus $T^2$ (identified here as $S^1 \times S^1$) and $D^2$ is homotopic to a circle. Therefore we can say that $A\cap B$ is path-connected and nonempty.
Since $X$ is the union of path-connected, closed subspaces whose intersection is path-connected and nonempty, we may apply Van Kampen's Theorem and compute $\pi_1 (X)$ to be $\pi_1 (X) \cong \big( \pi_1 (A) \ast \pi_1 (B) \big)/N$, where $N= \big\langle \phi_\ast (\alpha) \psi_\ast (\alpha)^{-1}: \alpha \in \pi_1 (A\cap B) \big\rangle$ and $\phi_\ast$ and $\psi_\ast$ represent the inclusion maps $$\phi_\ast : \pi_1 (A\cap B) \hookrightarrow \pi_1(A) \text{ and } \psi_\ast : \pi_1 (A\cap B) \hookrightarrow \pi_1 (B).$$
Since $\pi_1 (S^1 \times S^1) = \pi_1 (S^1) \times \pi_1 (S^1)$, we will have that $\pi_1 (A) \cong \Z \times \Z = \Z^2$. Next, it is clear that $\pi_1 (B)\cong 1$ since $B$ may be deformation retracted to a single point in the ambient space. Finally, since $A\cap B \simeq S^1$ we have that $\pi_1 (A\cap B) \cong \Z$.
Returning to the induced homomorphisms, since $\pi_1 (B)= 1$ we will have $\psi_\ast (\alpha)=1$ for any $\alpha \in \Z$ and also that $\phi_\ast (\alpha) = (\beta,\gamma) \in \Z^2$ for some $\beta,\gamma \in \Z$. Therefore we can simplify $N$ to the following presentation: $N = \big\langle \phi_\ast (\alpha) : \alpha \in \pi_1 (A\cap B) \big\rangle$. Moreover, if we define $\phi_\ast$ via $x\mapsto (x,1)\in \Z^2$ then $\phi_\ast$ is injective, so we know that $\im \phi_\ast \cong \Z$. Accordingly therefore $N\cong \Z$. Returning to the original computation, we get $\pi_1 (X) \cong \big( \pi_1 (A) \ast \pi_1 (B) \big)/N \cong (\Z^2 \ast 1)/N \cong \Z^2 / \Z \cong \Z.$ $\Box$
Is the proof correct?
I know the answer is right, but I also realized that $\mathbb Z$ can't embed into $\{1\}$, so $\psi_\ast$ is not injective, it would just be a constant map. I don't think this affects anything though?