Extending holomorphic $f$ on the unit disk to the boundary as $1/z$ I found the following problem statement in a book (Stein and Shakarchi): 

Show that there is no holomorphic function $f$ on the unit disk $\mathbb{D}$ that extends continuously to $\partial D$ such that $f\left(z\right) = 1/z $ for $z\in \partial D$

I know that I need uniform continuity of $f$ on $\mathbb{D}$ in order to show: 
$$\lim_{r\rightarrow 1^{-}} \int_{C_{r}}f = \int_{C_{1}} f = \int_{C_{1}} (1/z) = 2 \pi  i$$
And then I can get a contradiction. I think I remember $f$ being uniform continuous if it is continuous on a compact set - however I am not sure how to prove this. Also not sure how to prove the integral equations above given this fact. Any suggestions? 
 A: 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 $0<r<1$, you can show that $\int_{C_r}f(z)\,dz = r\int_{C_1}f(rz)\,dz$.  It follows that 
$$
\begin{align*}
\left|\int_{C_1}f(z)\,dz-\int_{C_r}f(z)\,dz\right|&=\left|\int_{C_1}f(z)-f(rz)\,dz+(1-r)\int_{C_1}f(rz)\,dz\right|\\
&\leq \left|\int_{C_1}f(z)-f(rz)\,dz\right|+(1-r)\left|\int_{C_1}f(rz)\,dz\right|\\
&\leq 2\pi\max\{|f(z)-f(rz)|:|z|=1\}+(1-r)2\pi M.
\end{align*}
$$
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.
A: You can prove this by contradiction. Suppose such an $f$ exists. Let $C_r$ be the circle of radius $r$ centered at the origin with positive orientation. By Cauchy's theorem $$\int_{C_r}f(z) dz=0$$ for all $0<r<1$. Since $f$ is continuous on $\partial D$, we can let $r\to 1$ to obtain $$\int_{C_1}f(z) dz=0.$$ But $$\int_{C_1}f(z)dz = \int_{C_1}\frac{1}{z}dz = 2\pi i \neq 0,$$ thus we have a contradiction.
