Can a disconnected set be simply connected?

The concept of "simply connected" really only makes sense for a path-connected space. That said, there are a few definitions that get thrown around, all of which are equivalent for path-connected spaces:

Definition 1 A simply connected space is a space in which every closed curve is homotopic to a point.

Definition 2 A simply connected space is a path-connected space in which every closed curve is homotopic to a point.

Definition 3 A simply connected space is a pointed space in which every path that starts and ends at the distinguished point is homotopic to a constant map via a homotopy through paths that all start and end at the distinguished point.

The definition used most often in algebraic topology (where this concept gets the most use) is Definition $3$, so I'd say this definition is the "correct" one. However, because the concept is so simple, in calculus or real analysis it's often cumbersome to have to introduce concepts like pointed space, so the definition is shortened to definition $1$ or $2$.

According to definition $1$, a disconnected space with simply connected components is simply connected. According to definition $2$, of course, it's not. According to definition $3$, the space has to have a distiguished point, and simple connectedness is determined solely by the path component containing that point. So under the actual definition, you could have one component be a violently non-simply-connected space like a Hawaiian earring, but as long as the component with the distinguished point is simply connected, the whole space is.

Allow me to reiterate that when we talk about simply connected spaces, in general we only want to worry about path-connected spaces, and the difference in all these definitions really boils down to how we want to handle the exceptions.

This is only an issue of terminology, but it's fair to say your professor is making a mistake about the terminology. The property that the disconnected set $B$ has doesn't have a very short name: it's that its fundamental group is trivial at every base point. That just means that every closed curve in $B$ can be contracted to a point, as your professor says. Call this property $P$. The reason that $P$ takes such a long time to name is that it's not important relative to being simply connected. A set is simply connected if it not only has $P$ but also admits a continuous path connecting any two points. Since $B$ is not path connected, we don't say it's simply connected even though it has property $P$.

For those who know some topology: the reason simple connectedness is more important than property $P$ is that it's saying $\pi_i$ is trivial for all $i\leq 1$. This generalizes immediately to the notion of $n$-connectedness, and $n$-connected spaces are fundamental in algebraic topology, for example in obstruction theory, the Hurewicz map between homotopy and homology, and the construction of Postnikov towers. Part of the significance is that such spaces are weakly equivalent to CW-complexes with one $0$-cell and no $1,2,...,n-1$-cells.