Does the interior of a closed jordan curve in the plane depend continuously on the curve? Given a continuous map 
$$\gamma \colon S^1 \times [0,T] \to \mathbb{R}^2$$ 
where the map $s\mapsto \gamma(s,t)$ is an injection of $S^1$ into the plane (i.e. a Jordan curve) for each $t\in [0,T]$, by the Jordan curve theorem it follows that for each $t$ the curve $s\mapsto \gamma(s,t)$ divides the plane into a pair of disjoint connected open sets consisting of a non-compact exterior region and a compact interior region
$$A(t)\subset\mathbb{R}^2.$$ 
I am interested in the continuous dependence of $A(t)$ on $t\in [0,T]$ in the following sense. Defining for each $t\in[0,T]$ the continuous distance function 
$$d(\cdot,t)\colon \mathbb{R}^2 \to \mathbb{R}$$ 
by $d(p,t)=\mathrm{dist}(p,A(t)^c)$, so that $d(p,t)>0$ for $p\in A(t)$ and $d(p,t)=0$ for $p\in A(t)^c$, is $d$ continuous in $t$?
NOTE: I would also be interested to see proofs which hold for a smooth flow of curves i.e. a smooth map $\gamma\colon S^1\times [0,T] \to \mathbb{R}^2$ for which $s\mapsto \gamma(s,t)$ is an immersion for all $t\in [0,T]$, if no proof can be given for the more general case!
 A: $d(p,t)$ is indeed continuous in $t$. 
To see why, first note that $d(p,t) = \operatorname{dist}(p,\operatorname{image}(\gamma_t))$ (where $\gamma_t(s)=\gamma(s,t)$), because the line segment from $p$ to any point of $A(t)^c$ must pass through $\operatorname{image}(\gamma_t)$. 
Second, note that the function $t \mapsto \gamma(\cdot,t)$ is continuous in the compact open topology on functions $S^1 \mapsto \mathbb R^2$ (a standard topology exercise). And since the circle $S^1$ is compact, one can prove furthermore that the map $t \mapsto \operatorname{image}(\gamma_t)$ is continuous in the Hausdorff metric, which is the metric on compact subsets of $\mathbb R^2$ defined by 
$$d_{\text{Haus}}(A,B) = \operatorname{inf}\{r \ge 0 \mid A \subset N_r(B), B \subset N_r(A)\}
$$
In particular, $d(p,t)$ is equal to $d(p,q)$ for $q \in \operatorname{image}(\gamma_t)$. For any $\epsilon > 0$ there exists $\delta > 0$ such that if $|t-t'|<\delta$ then $d_{\text{Haus}}(\operatorname{image}(\gamma_t),\operatorname{image}\gamma_{t'}) < \epsilon$ and so $\operatorname{image}(\gamma_{t'})$ contains a point $q'$ such that $d(q,q') < \epsilon$, hence $|d(p,t) - d(p,t')| < \epsilon$.
With a little more work, one can even show that $d(p,t)$ is jointly continuous in $p$ and $t$, i.e. it is a continuous function on the subset of $\mathbb R^2 \times [0,1]$ consisting of all $(p,t)$ such that $p \in A(t)$.
