In this case, you will actually need two base steps, since your recursion involves two previous terms. Also, note that your closed form should be $$a_n=\frac23\left(1+\frac{(-1)^{n+1}}{2^n}\right).\tag{$\star$}$$ Our two base cases (which I leave entirely to you) will be $n=0$ and $n=1.$
Now, for the induction step we must assume that it holds for the previous two cases--that is, suppose we have an integer $k\ge 2$ such that $(\star)$ holds for $n=k-2$ and $n=k-1.$ We must show that it holds for $n=k$.
Since $k\ge 2,$ then $$a_k=\frac12\left(a_{k-2}+a_{k-1}\right)$$ by definition, and since $(\star)$ holds for $n=k-2$ and $n=k-1,$ we then have $$a_k=\frac12\left(\frac23\left(1+\frac{(-1)^{(k-2)+1}}{2^{k-2}}\right)+\frac23\left(1+\frac{(-1)^{(k-1)+1}}{2^{k-1}}\right)\right).$$ After some arithmetic manipulations (which I leave to you), we will indeed see that $$a_k=\frac23\left(1+\frac{(-1)^{k+1}}{2^k}\right),$$ as desired.
In general, if our recursion involves multiple previous terms, then we will need multiple base cases and multiple induction hypotheses. For (a monumentally contrived) example, suppose that $b_0=0,b_1=1,b_2=2,b_3=3,b_4=4,b_5=5$ and that for $n\ge 6,$ we have $$b_n=b_{n-5}+\frac53\left(b_{n-3}-b_{n-6}\right).$$ I claim that $b_n=n$ for all $n$. Now, since the recursion doesn't start until term $6,$ then we will need $6$ base cases ($n=0$ through $n=5,$ all of which are immediate in this example). The recursion, itself, involves terms ranging from $3$ prior (at the least) through $6$ prior (at the most), so we need $6-3+1=4$ assumptions in our inductive hypothesis. In particular, we need to assume that there is some $k\ge6$ such that $b_n=n$ for $n=k-6,k-5,k-4,k-3.$ We then want to show that $b_n=n$ when $n=k$ (and possibly justify that this completes the proof). Indeed, $$b_k=b_{k-5}+\frac53\left(b_{k-3}-b_{k-6}\right)=(k-5)+\frac53\bigl((k-3)-(k-6)\bigr)=k-5+\frac53\cdot 3=k.$$ Now, since $b_0,b_1,b_3$ have the desired form, then $b_6$ is taken care of; since $b_1,b_2,b_4$ have the desired form, then $b_7$ is taken care of; since $b_2,b_3,b_5$ have the desired form, then $b_8$ is taken care of; etc. In this fashion, we see that $b_n=n$ for all integers $n\ge 0.$
You can generalize this idea to prove a closed form for any recursively defined sequence.