You’ll have a hard time proving that $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right] = \left[-1, \frac1n\right]\;:$ it doesn’t even make sense. The $n$ on the lefthand side is a dummy variable: the value of the expression wouldn't change if you replaced it by something else, say $k$, to get $\bigcup_{k = 1}^\infty \left[-1, \frac1k\right]\;.$ The $n$ on the righthand side, however, is apparently a particular integer. Thus, you’re using one letter, $n$, to represent two unrelated things of very different kinds.
You’re actually being asked to figure out exactly what real numbers are in the set $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$ and then to express the set in a way that doesn’t require talking about infinitely many sets. For example, it should be clear that every real number in the interval $[-1,0]$ belongs to the set. However, $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]\ne[-1,0]\;,$ because, for instance, $\frac13\in\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]\;,$ but $\frac13\notin[-1,0]$.
Sketch the intervals $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$ for the first few positive integers $n$ to get an idea of what they look like. Once you’ve done that, figure out exactly what the set $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$ looks like, and write down a simple description of that set. Finally, if your set is $A$, prove that $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]=A\;,$ probably by showing that if $x\in A$, then $x\in\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$, and if$x\in\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$, then $x\in A$.