I have two questions regarding the following exercise. First, is my solution (see below) correct? Second, how would the solution go that the writer of the book had in mind (i.e. using the hint)?
Let $K\subset\mathbb{C}$ be a compact set such that $\mathbb{C}-K$ is connected and such that for any $\epsilon>0$ there exist balls with $K\subset\bigcup_jB(a_j,r_j)$ and $\sum_jr_j<\epsilon$; you may assume (for this exercise) that the balls can be chosen disjoint for every $\epsilon>0$. Let $U\supset K$ be open. Proof that every holomorphic function on $\mathbb{C}-K$, bounded on $U-K$ can be analytically continued to $\mathbb{C}$.
It is given as a hint to use Cauchy's theorem, but I had no clue how Cauchy's theorem is of any help here; I came up with a solution myself.
So take an $a\in K$, then for every $\epsilon>0$ we can find an $\alpha_{\epsilon}\in K$ and an $0<r_{\epsilon}<\epsilon$ such that $a\in B(\alpha_{\epsilon},r_{\epsilon})$; choosing the balls disjoint we could find $\delta\in(0,\epsilon)$ such that $A(\alpha_{\epsilon},r_{\epsilon}+\frac{\delta_{\epsilon}}{2},r_{\epsilon}+\delta_{\epsilon})\subset\mathbb{C}-K$. Now let's choose $\epsilon$ small enough such that the $\epsilon$-neighbourhood of $K$ is contained in $U$; than by the boundedness property there exists an $M$ such that for any such $\epsilon$ $f$ is bounded by this $M$ on $A(\alpha_{\epsilon},r_{\epsilon}+\frac{\delta_{\epsilon}}{2},r_{\epsilon}+\delta_{\epsilon})$. But if $\epsilon\to0,r_{\epsilon}\to0$ and $\alpha_{\epsilon}\to a$. Now take a sequence $(\epsilon_ n)_{n\in\mathbb{N}}$ converging to $0$, then we can for any $n\in\mathbb{N}$ freely choose a $\beta_n\in A(\alpha_{\epsilon_n},r_{\epsilon_n}+\frac{\delta_{\epsilon_n}}{2},r_{\epsilon_n}+\delta_{\epsilon_n})$, and by boundedness we would have $\lim_{n\to\infty}(z-a)f(\beta_n)=0$; because this is true for any such sequence $(\epsilon_ n)_{n\in\mathbb{N}}$ converging to $0$ and any choice for the $\beta_n$'s, we have $\lim_{z\to a}(z-a)f(z)=0$. Thus $f$ has a removable singularity in $a$. The $a\in K$ was arbitrary; for any such $a$ we remove the singularity; we get an analytic continuation $\tilde{f}$ on $\mathbb{C}$.
First note $((z-1)(z+1))^{1/2}$ is analytic on $\Bbb{C}- [-1,1]$, it is bounded on $|z| \le 2$, but it is not analytic on $\Bbb{C}$. So $K = [-1,1]$ doesn't work.
If $f$ is analytic on $\Bbb{C} - K$ then for any curve $\gamma$ enclosing $K$ counter clockwise and $K \subset B_0(r)$ and $|z| < R, z \not \in K$ $$f(z) = \frac{1}{2i\pi} \int_{|z|= R} \frac{f(s)}{s-z}ds - \frac{1}{2i\pi} \int_\gamma \frac{f(s)}{s-z}ds$$
Your assumption is that $K \subset B_0(R)$, $\sup_{|z| \le R} |f(z)| \le C$ and for every $\epsilon$, there are $a_{j,\epsilon},r_{j,\epsilon}$ such that $K \subset \bigcup_{j=1}^{J_\epsilon} B(a_{j,\epsilon},r_{j,\epsilon})$ and $\sum_{j=1}^{J_\epsilon} r_{j,\epsilon} < \epsilon$.
Thus $\gamma_\epsilon = \bigcup_{j=1}^{J_\epsilon} \partial B(a_{j,\epsilon},r_{j,\epsilon})$ encloses $K$ and $$| \int_{\gamma_\epsilon} \frac{f(s)}{s-z}ds| = |\sum_{j=1}^{J_\epsilon} \int_{\partial B(a_{j,\epsilon},r_{j,\epsilon})}\frac{f(s)}{s-z}ds| \le \sum_{j=1}^{J_\epsilon}C 2\pi r_{j,\epsilon} \le C2\pi \epsilon $$ Whence $\int_\gamma \frac{f(s)}{s-z}ds =\lim_{\epsilon \to 0} \int_{\gamma_\epsilon} \frac{f(s)}{s-z}ds= 0$ ie. $$f(z) = \frac{1}{2i\pi} \int_{|z|= R} \frac{f(s)}{s-z}ds = \frac{1}{2i\pi} \int_{|z|= R} \frac{f(s)}{s}\sum_{k=0}^\infty (z/s)^k ds=\sum_{k=0}^\infty z^k \frac{1}{2i\pi} \int_{|z|= R} \frac{f(s)}{s^{k+1}}ds$$ is analytic and entire.