After thinking about it for a while, I believe that the following space is not simply connected, but don't know how to prove it (either way).
Let $Q = [0, 1]^2$ be the unit square. The space is
$$X = \{ (z_1, z_2) \in Q^2 ~:~ z_1 \neq z_2 \},$$
which is like a $4$-dimensional cube minus a $2$-dimensional "diagonal".
Question: How do you prove that $X$ is or isn't simply connected?
There are some details to be filled in here, but here goes.
You can rotate your $X$ so that your "plane $z_1=z_2$" becomes a coordinate $2$-plane, say $P=\{(0,0,x_3,x_4):x_3,x_4\in\Bbb R\}$. Then $X$ is mapped to a homeomorphic $X'$ with $X'\subseteq \Bbb R^4-P$. But $\Bbb R^4-P$ is homotopy equivalent to a $\Bbb R^2-O$ and so to the circle $S^1$ which has fundamental group $\Bbb Z$.
Idea: take a loop in $\Bbb R^4-P$ which does not contract to a point in that set, and manoeuvre it so that it lies inside $X'$. That will show that $\pi_1(X)\cong \pi_1(X')\ne0$.