Find an open connected set $G\subseteq\mathbb{C}$ and two continuous functions $f,g$ defined on $G$ such that $f(z)^2=g(z)^2=1-z^2$. Can you make $G$ maximal? Are $f$ and $g$ analytic?
The following holds: $\log f(z)^2= \log (1-z^2)$. Here $\log: C-\{z\in \mathbb R: z\le 0\} \to C$ is the branch of the logarithm.
Let $G$ not contain any $z\in \mathbb C$ such that $\Re(1-z^2)\le 0\implies G\supset\{z\in \mathbb C:\Re(z^2)< 1\}$.
But I'm not sure how to go from here.
For this kind of problems, the solution is not unique in general. For example, if you want to define $\sqrt{z}$, any simply connected domain that doesn't contain $0$ fits like $\mathbb{C}\backslash\lambda\mathbb{R_+}$ for any $\lambda \in \mathbb{S}^1$ or even the plane minus a spiral that have zero for origin.
You idea of using $\log$ is the good one. You want to be able to define $f(z) = e^{\frac{\log(1 - z^2)}{2}}$ (which is holomorphic hence continuous) so you only need $\log(1 - z^2)$ to be define. If you take the classical domain of the complex $\log$ : $\mathbb{C}\backslash\mathbb{R}_-$, you can define $f$ on the set $G = \{z \in \mathbb{C}|1 - z^2 \notin \mathbb{R}_-\} = \mathbb{C}\backslash(]-\infty,-1] \cup [1,+\infty[)$, which is open and connected, but it is not the only possibility !
Notice that $G$ strictly contains $\{z \in \mathbb{C}|\Re(1 - z^2) > 0\}$. And actually, $G$ is maximal. Indeed, you could continue $f$ at $\pm 1$ with $f(-1) = f(1) = 0$ (just like the square root can be continued at $0$) but in this case, $G$ would be no more open and $f$ no more holomorphic.
Assume that $f$ can be continued on some open domain $H$ containing $G$ strictly. By openness of $H$, it contains at least one point $z_0 > 1$ or $z_0 < -1$. Assume that $z_0 > 1$ without loss of generality.
However, when $z$ is close to $z_0$ and $\Im(z) > 0$, then $\Im(1 - z^2) = -\Im(z^2) = -2\Re(z)\Im(z) < 0$ so the argument of $1 - z^2$, chosen in $]-\pi,\pi[$, approaches $-\pi$, which means that $\log(1 - z^2) \rightarrow \ln(z_0^2 - 1) - i\pi$ when $z \rightarrow z_0$ and $\Im(z) > 0$. Similarly, when $z \rightarrow z_0$ with $\Im(z) < 0$, $\log(1 - z^2) \rightarrow \ln(z_0^2 - 1) + i\pi$.
It implies that $f(z) \underset{z \rightarrow z_0, \Im(z) > 0}{\longrightarrow} -i\sqrt{z_0^2 - 1}$ but $f(z) \underset{z \rightarrow z_0, \Im(z) < 0}{\longrightarrow} i\sqrt{z_0^2 - 1}$, which implies that $f$ can't be continuously continued at $z_0$.
