= Solution
{c}
The union is bounded but is not a <closed set>: the points $-1+1/n$, $n\geq2$, belong to the open disc and tend to $-1$, which lies in neither disc. Thus it is \b[not compact] by the <Heine-Borel theorem>.
Nevertheless it is \b[connected], indeed <path-connected>. Every point of the open disc joins $0$ by a straight segment. The segment from $0$ to $1$ stays in the union, since its points before the endpoint belong to the open disc and the endpoint $1$ belongs to the closed disc. Every point of that closed disc joins $1$ inside it by <convexity>. Concatenating these paths joins any two points of the union. The fact that the two discs are disjoint does not imply disconnection: their contact point belongs to one of them. This illustrates <connectedness at a one-sided contact point>.
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