I have two related problems,

Problem 1: If $f$ is continuous on $B(0,2)$ and holomorphic on $B(0,2)\backslash [0,1]$ show that $f$ is actually holomorphic on $(0,2)$.

We can use $\Gamma = \{|\zeta| = r \}- \alpha_\varepsilon$ where $\alpha_\varepsilon = \varepsilon \cos t/2 + i \varepsilon \sin t$. The curve $\alpha_\varepsilon$ should be a circle centered at $x=1/2$ which squeezes down on the interval $[0,1]$.

Then Ind$(\Gamma , z) = 1$ for $z \in B(0,2)\backslash[0,1]$ and $\Gamma \approx 0$ in $B(0,2)\backslash[0,1]$. The global Cauchy formula then gives $$f(z) = \frac{1}{2\pi i} \int_{|\zeta| = r} \frac{f(\zeta)}{\zeta - z} d\zeta - \frac{1}{2 \pi i} \int_{\alpha_\varepsilon} \frac{f(\zeta)}{\zeta-z} d\zeta.$$

Because $\frac{f(\zeta)}{\zeta -z}$ is continuous it's possible to show that $\lim_{\varepsilon \to 0} \frac{1}{2 \pi i} \int_{\alpha_\varepsilon} \frac{f(\zeta)}{\zeta-z} d\zeta = 0$ using uniform continuity and an appropriate estimate.

The formula $\frac{1}{2\pi i} \int_{|\zeta| = r} \frac{f(\zeta)}{\zeta - z} d\zeta$ should define a function which is holomorphic on the whole disk.

Problem 2: Give an example of a function $f$ which is holomorphic on $B(0,2)\backslash [0,1]$ and bounded on $B(0,2)\backslash[0,1]$ which cannot be extended to be holomorphic on $B(0,2).$

OK, now I am confused. I'm not sure if I should just try to use $$f(z) = \frac{1}{2\pi i} \int_{|\zeta| = r} \frac{f(\zeta)}{\zeta - z} d\zeta - \frac{1}{2 \pi i} \int_{\alpha_\varepsilon} \frac{f(\zeta)}{\zeta-z} d\zeta$$ or if I should try to find a function which is discontinuous on the line segment defined by $[0,1]$ but also bounded. Any guesses/advice?