Closed circle as a metric space

Yes, $(C,d)$ is a metric space. And, yes, for points in $C$, the open balls are intersection of usual open balls (in $\mathbb{R}^2$) with $C$.

Every metric space $X$ is both open and closed in itself; note that $\varnothing$ is by definition both closed and open and that $\varnothing = X^{c}$. So $C$ is open in itself but, when considered as a subset of $\mathbb{R}^{2}$, is closed in $\mathbb{R}^{2}$.

I assume by a circle you really mean a circle rather than a disk. Let $a \in C$ and $\delta > 0$. Note that $B(a,\delta) := \{ x \in \mathbb{R}^{2} \mid |x-a| < \delta \}$ is an open ball that looks like an open disk. The set $\{ x \in C \mid |x-a| < \delta \} = B(a,\delta) \cap C$ is an open ball in $C$ but looks not like a ball.

If $(X,d)$ is a metric space and $A\subseteq X$, then $(A,d')$ is a metric space, where $d'$ is just $d$ restricted to $A$; namely, $d':=d|_{A\times A}$. Usually $d$ will be written instead of $d'$, and we'll just understand by context that it's the restriction. Now $A$ is open in the $d'$ metric, regardless whether or not it's open in the $d$ metric.

A different point of view is your (ii). If $x\in A$, then the $d$'-ball centered at $x$ with radius $r>0$ is equal to the intersection of $A$ with the $d$-ball centered at $x$ with radius $r$. In symbols,

$$ B_{d'}(x,r) = A\cap B_d(x,r). $$ There is a partial converse to this statement as well. If $B_d(x,r)$ is a $d$-ball with $x\in X$, then the intersection $A\cap B_d(x,r)$ is open in the $d'$ metric. Indeed, if $a \in A\cap B_d(x,r)$, then setting $\delta:=r-d(x,a)$ implies $B_{d'}(a,\delta)\subseteq A\cap B_d(x,r)$. To see this, suppose $y\in B_{d'}(a,\delta)$. Then $y\in A$ by the fact that $d'$ only acts on pairs in $A$, and $y\in B_{d}(x,r)$ because $d(y,x) \leq d(y,a)+d(a,x) < \delta+d(a,x)=r$.