A simple fundamental group I think I have this one, but I want to make sure:

Using Van Kampen's theorem,

Find the fundamental group of two disjoint spheres with each north pole identified, and each south pole identified.

The open sets I chose are sphere 1 minus the north pole, sphere 2 minus the south pole. The intersection is nonempty with trivial fundamental group since poles are identified, so we get trivial*trivial=trivial. Correct?
Let $X$ be your space. Another possibility, not using van Kampen theorem (explicitely), is to glue a 3-cell inside each sphere. Because the fundamental group depends only on the 2-skeleton, $\pi_1(X)$ is isomorphic to the fundamental group of our new space $\tilde{X}$. Now, $\tilde{X}$ clearly retracts by deformation on a circle, hence $$\pi_1(X) \simeq \pi_1(\tilde{X}) \simeq \pi_1(\mathbb{S}^1) \simeq \mathbb{Z}.$$
First of all, we may picture the two spheres as a torus pinched in two points, namely North and South points of the two spheres.
Doing this, it's clear that our space $X$ is path connected and can be covered by two open sets: $A = \left[\left(S^2 \amalg S^2 \right)/ \sim \right] \setminus \left[\,\{N\}\,\right]$ where $\sim$ is the relation with North and South poles identified and $\left[\,\{N\}\,\right]$ is the class of equivalence of North poles of the two spheres; and $B$, the first half of the 2-pinched torus, namely $\left[ S^2 \amalg \{S\} \right]$.

$A$ and $B$ are obviously open sets and path connected; $A$ is homeomorphic to $S^2 \vee S^2$, $B$ is homeomorphic to $S^2$. Their intersection $A\cap B$ is homeomorphic to an infinite cylinder and thus its fundamental group is $\mathbb{Z}$.

Now, Van Kampen says that $\pi(X) = \pi(A) \times \pi(B)$ if $A \cap B$ is simply connected. In a more general version, it says that $\pi(X) = \left[ \pi(A) \times \pi(B) \right] \star \pi(A\cap B)$, where $\star$ is the free product. Now we know that $\pi(A)$ is trivial (again for Van Kampen, if you desire) and $\pi(B)$ is trivial. So, $\pi(X) = \pi(A\cap B) = \mathbb{Z}$.