Is $\mathbb{R}^4$ minus a line simply connected?

You have $\mathbb{R}^4\setminus \{(0,0,0,w) | w\in \mathbb{R} \} = (\mathbb{R}^3\setminus \{(0,0,0 \}) \times \mathbb{R}$, thus $\mathbb{R}^4\setminus \{(0,0,0,w) | w\in \mathbb{R} \}$ is homotopy equivalent to $\mathbb{R}^3\setminus \{(0,0,0 \}$. This space is homotopy equivalent to $S^2$ (in fact, $S^2$ is a strong deformation retract of $\mathbb{R}^3\setminus \{(0,0,0 \}$) which is known to be simply connected.

A way to visualize curves in $\mathbb{R}^4$ is to think about the curves as being in $\mathbb{R}^3$ through a projection, then keeping track of the $w$-coordinate through color along the curve. Two curves can pass through each other without intersecting in $\mathbb{R}^4$ if they have different colors when they intersect in $\mathbb{R}^3$.

In the projection onto the $xyw$ space, $\mathbb{R}^4$ minus a line is $\mathbb{R}^3$ minus a line. If you have a closed loop in the complement, you can unhook it from the line by changing the color of the loop so that it is a distinct color from the line.

In the projection onto the $xyz$ space, it's $\mathbb{R}^3$ minus the origin, in which case it's fairly obvious there's no obstruction to deforming any loop to a point!

Yet another way to see it is to deformation retract $\mathbb{R}^4-\{(0,0,0,w):w\in\mathbb{R}\}$ to the set of unit vectors in this space. This results in $S^3$ minus two points, which is simply connected.