As the real and the imaginary part of $\phi$ are continuous on the compact interval $I$ the point $\infty$ is not part of the problem.
The essential ingredient in the following proof is the compactness of $I$ resp. its image in $D$.
For $t\in I$ put $\rho(t):=\sup\{r>0\ |\ D_r\bigl(\phi(t)\bigr)\subset D\}$. We may assume $\rho(t)<\infty$ for all $t\in I$, or there would be nothing to prove. The function $t\mapsto \rho(t)$ is positive and continuous on $I$. To prove the latter assume that an $\epsilon>0$ is given. Then there is a $\delta>0$ with $|\phi(t)-\phi(t')|<{\epsilon\over 2}$ as soon as $|t-t'|<\delta$. It follows that $\rho(t')\geq \rho(t)-{\epsilon\over 2}$ when $|t-t'|<\delta$, whence by symmetry $|\rho(t')- \rho(t)|\leq{\epsilon\over 2}$ when $|t-t'|<\delta$.
Therefore we may conclude that there is a $\delta>0$ with $\rho(t)>2\delta$ for all $t\in I$. Choose $N$ so large that $|\phi(t)-\phi(t')|<\delta$ when $|t-t'|, and put $t_j:=T_0+ jh$ $\ (0\leq j\leq N)$. Then the open disks $\Delta_j$ $\ (1\leq j\leq N)$ with center $\phi(t_j)$ and radius $\delta$ are subsets of $D$ and contain the respective arcs $\phi\bigl([t_{j-1},t_j]\bigr)$.