Let $D:=\{z\in \mathbb{C}: \lvert z \rvert < 1\}$ and $B := \{z\in \mathbb{C}: \lvert z \rvert = 1\}$. Let $f:D\to \mathbb{C}$ be a uniformly continuous function. Then $\lim\limits_{z\to z_0, z\in D} f(z)$ exists for all $z_0\in B$.
I'm not sure how to approach this problem. Here's my guess: since points in $B$ are limit points of $D$, $\exists$ a sequence $\{z_n\}$ converging to $z_0 \in B$, for any $z_0\in B$. Now, since $f$ is continuous, $\lim\limits_{n\to \infty} f(z_n) = \lim\limits_{z\to z_0, z\in D} f(z)$ must exist. But if the proof were this straightforward, one could do away with simple continuity, but in this case uniform continuity is implied. So I think that I'm missing something.
The problem seems to be the distinction between continuity and continuity at the boundary. Since the function is originally not defined at the boundary, you should not apply the definition of continuity at the boundary.
When you have a sequence converging to a point on the boundary, the sequence itself is Cauchy sequence. Uniform continuity gives you the image of the sequence being cauchy.