First it would be helpful if you use different notation for the original quantities and their normalized versions (i.e., what you refer to as proportions). I am going to use $a,b$ to stand for the normalized version of $A,B$ respectively.
Secondly, in terms of statistical test, your null hypothesis seems to be that $a=b$, in other words, $a=0.5$. One of the crucial assumptions of t-test is that the sample should follow a normal distribution. Given that $a$ is bounded between $0$ and $1$, this is patently false.
Instead, let's make the follow assumptions: $A_i$, $B_i$, and therefore $a_i$, $b_i$ are independent identically distributed. The rest is all mathematically rigorous. You can forget about $A_i,B_i$ and focus only on $a_i,b_i$.
If your sample size $N$ is large enough, central limit theorem implies that $\sum_{i=1}^N a_i / \sqrt{N}$ is close in distribution to the normal, provided $a_i$ are independent, which is true in your case. Furthermore the boundedness of $a_i$ actually helps you here because you can get uniform bound on the convergence rate, in terms of metrics like Levy's metric or Wasserstein's metric; the rate of convergence depends on first three moments, by the Berry-Esseen theorem. With that you can estimate the probability that $\frac{1}{N} \sum a_i = \bar{a}$ given $\mathbb{E}(a_i) \equiv 0.5$, pretending that $\frac{1}{\sqrt{N}} \sum_{i=1}^N a_i$ follows the normal distribution, with a tiny error term coming from CLT.
Now given that your sample size is pretty small, and very little is known about the distribution of $A_i,B_i$ and therefore $a_i,b_i$, it is always a good idea to err on the side of caution. Let's say the sample mean of $a_i$ you observed was $\bar{a}$. So you could try to compute the following quantity:
$$ \sup_{\mathcal{L}(A,B)} \mathbb{P}(\mid \frac{1}{5} \sum_{i=1}^5 a_i - 0.5 \mid > \mid \bar{a} -0.5\mid \big\rvert \mathbb{E}(a_i) \equiv 0.5).$$
