A family that is not equicontinuous in the unit disc 
*

*How to show that if  a family of (complex-valued) functions in the unit disc $D$ is equicontinuous  in every compact subset of $D$
then it is equicontinuous
in $D$?

*What is an example of a family of (complex-valued) functions in the unit disc $D$ that is not uniformly equicontinuous
in $D$ but uniformly  equicontinuous  in every compact subset of $D$?

 A: There is no such example -- in other words, if a set of functions is equicontinuous on every compact subset of $D$, then it is equicontinuous on $D$.
Proof. Let $F$ be a set of functions $D \to \mathbb{C}$ such that $F$ is equicontinuous on every compact subset of $D$. Now let $x \in D$ be arbitrary. Let $t > 0$ be such that $\overline{B_t(x)} \subseteq D$. Since $\overline{B_t(x)}$ is compact, $F$ is equicontinuous on $\overline{B_t(x)}$.
In particular, for all $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $f \in F$ and all $y \in \overline{B_t(x)} \cap B_\delta(x)$, we have $\lvert f(y) - f(x) \rvert < \varepsilon$. Without loss of generality, we may suppose $\delta < t$ so that $\overline{B_t(x)} \cap B_\delta(x) = B_\delta(x) = D \cap B_\delta(x)$. But now we have shown precisely that $F$ is equicontinuous at $x \in D$! Since $x$ was arbitrary, we conclude that $F$ is equicontinuous on $D$. $\square$

Here's an alternate question: is there a set of functions which is uniformly equicontinuous on every compact subset of $D$, but not uniformly equicontinuous on $D$? Now the answer is "yes", but not for a very interesting reason.
Let $f : D \to \mathbb{C}$ be the function $f(z) = (1-\lvert z \rvert)^{-2}$. This function is continuous but not uniformly continuous (exercise!). Thus, $F = \{f\}$ is equicontinuous but not uniformly equicontinuous. However, continuity implies uniform continuity on compact domains. Thus, $F$ is uniformly equicontinuous on all compact subsets of $D$. This is an "uninteresting" example because it doesn't highlight anything interesting about equicontinuity: the set $F$ was a singleton, so (uniform) equicontinuity of $F$ is the same as (uniform) continuity of $f$!
