The argument in your first sentence is correct. But you need to justify the statement "obviously $A$ is unbounded". Why is this so? Well, of course, because $A$ contains all even powers of $2$ and the set of all even powers of 2 is unbounded from above. So, what you need to show is that:
given any positive integer $n$, there is an even positive integer $k$ so that $2^{ k}\ge n$.
The above claim would follow from a more general statement:
given any positive integer $n$, the inequality $2^n\ge n$ holds.
You can prove this statement by induction.
So, I would start of by proving the Lemma: $2^n\ge n$ for any positive integer $n$. Then to show $A$ is unbounded, argue as follows:
Let $n$ be a positive integer.
If $n$ is even, then $2^n=(-2)^n$ is in $A$ and from the lemma we have $(-2)^n=2^n\ge n$.
If $n$ is odd, we have $2^{n+1}=(-2)^{n+1}$ is in $A$ and, again from the lemma, $(-2)^{n+1}=2^{n+1}\ge n+1>n$.
So in either case, we have an element in $A$ greater than or equal to $n$. Since $n$ was an arbitrary positive integer, it follows that $A$ is not bounded from above.
Note that, to show that your $A$ is not bounded, you need only show that $A$ is not bounded from above (or show that it is not bounded from below). Of course, if you were required to show that $A$ is unbounded in both directions, you'd use a similar argument to show that $A$ is not bounded from below.