Yes, a continuous function on a compact set is uniformly continuous. "...however I am not sure how to prove this." Are you looking to prove this standard result just to apply it in one case, or would you be comfortable just citing it for now? Here's a reference.
Suppose $f$ has a continuous extension to the boundary. That means the extension, which we're still calling $f$, is continuous on the closed disk, which is a compact set because it is closed and bounded. This implies $f$ is uniformly continuous, by what is apparently called the Heine–Cantor theorem. It also implies that $f$ is bounded, i.e., there exists some $M>0$ such that $|f(z)|\leq M$ for $|z|\leq 1$.
Given $r$ with $0
$$
\begin{align*}
\left|\int_{C_1}f(z)\,dz-\int_{C_r}f(z)\,dz\right|&=\left|\int_{C_1}f(z)-f(rz)\,dz+(1-r)\int_{C_1}f(rz)\,dz\right|\\
&\leq \left|\int_{C_1}f(z)-f(rz)\,dz\right|+(1-r)\left|\int_{C_1}f(rz)\,dz\right|\\
&\leq 2\pi\max\{|f(z)-f(rz)|:|z|=1\}+(1-r)2\pi M.
\end{align*}
$$
By uniform continuity, given $\varepsilon>0$ there exists $\delta>0$ such that $|w-z|<\delta$ implies $|f(w)-f(z)|<\frac{\varepsilon}{4\pi}$. If $r$ is chosen such that $1-r<\delta$, and such that $1-r<\dfrac{\varepsilon}{4\pi M}$, or in other words, $r>\max\left\{1-\delta,1-\frac{\varepsilon}{4\pi M}\right\}$, then for all $z$ with $|z|=1$ we have $|f(z)-f(rz)|<\frac{\varepsilon}{4\pi}$ by choice of $\delta$, and therefore
$$\left|\int_{C_1}f(z)\,dz-\int_{C_r}f(z)\,dz\right|< 2\pi \frac{\varepsilon}{4\pi} + \dfrac{\varepsilon}{4\pi M}2\pi M=\varepsilon.$$
This shows that $\lim\limits_{r\nearrow 1}\int_{C_r}f(z)\,dz = \int_{C_1}f(z)\,dz$. All that was used was that $f$ is continuous on the closed disk. (Incidentally, in your particular case you would have $M=1$ by the maximum modulus theorem.) This equality would provide a contradiction in your case as you mentioned, because it implies that $\int_{C_1}f(z)\,dz = 0$ for any $f$ that is the continuous extension of a holomorphic function on the open disk, by Cauchy's theorem on the disk.
