Different conditions for Continuity and Uniform Continuity of a function on Union of Sets My textbook defines the following, respectively, for continuity and uniform continuity on union of sets: 
Continuity : 

Where $\operatorname{Cl}$ refers to the closure of the respective set.

Uniform Continuity:

where $\operatorname{dist} (A,B) = \inf \{d(a,b)\mid a \in A, b \in B\}$


The book further goes on to add that for uniform continuity, the condition $\operatorname{dist}(A,B) > 0$ cannot be weakened to $\bar A \bigcap \bar B = \varnothing$

Could someone please explain to me why is there this discrepancy between the conditions? Why does one require positive distance from the closure of other sets for continuity while uniform continuity requires  positive distance from other set only ?
 A: 
Why does one require positive distance from the closure of other sets for continuity

You are misunderstanding the condition.The condition says that every member of $\mathcal{C}$ shall not intersect the closure of the union of the other members, that is much weaker than having a positive distance from (the closure of) that union. This condition is for example satisfied if each member of $\mathcal{C}$ is open, regardless of the distance between the sets. For example, in $\mathbb{R}$ (with the standard topology, induced by $d(x,y) = \lvert x - y\rvert$ or by $d(x,y) = \lvert \arctan x - \arctan y\rvert$ or a lot of other metrics) the collection $\mathcal{C} = \{(n, n+1) : n \in \mathbb{Z}\}$ satisfies the condition, and the distance of each member of $\mathcal{C}$ to the union of the other members is $0$.
If you understand the terminology: Setting $Z = \bigcup \mathcal{C}$ and endowing it with the subspace topology, induced by the restriction $d\lvert_{Z \times Z}$ of the metric on $X$, the condition says that each $A \in \mathcal{C}$ shall be open in $Z$ (which automatically is the case if $A$ is open in $X$).
However, since $B = \operatorname{Cl}\bigl(\bigcup(\mathcal{C}\setminus \{A\})\bigr)$ is a closed set, every point $x \in X\setminus B$ has a positive distance from $B$. And since the condition says $A \cap B = \varnothing$, every $a \in A$ has positive distance from $B$.

while uniform continuity requires positive distance from other set only

Since in that case $\mathcal{C}$ is assumed to be finite, a positive distance from each other member means a positive distance from the union of the other members. Let's say $\mathcal{C} = \{A_1, \dotsc , A_n\}$, for $i \neq j$ define
$$\eta_{ij} = \operatorname{dist}(A_i,A_j)$$
and for each $i$ ($1 \leqslant i \leqslant n$) set $\delta_i = \min \{ \eta_{ij} : 1 \leqslant j \leqslant n, j \neq i\}$. Then
$$0 < \delta_i = \operatorname{dist}\biggl(A_i, \bigcup \bigl(\mathcal{C}\setminus \{A_i\}\bigr)\biggr) = \operatorname{dist}\biggl(A_i, \operatorname{Cl}\Bigl(\bigcup \bigl(\mathcal{C}\setminus \{A_i\}\bigr)\Bigr)\biggr)$$
for each $i$, and finally
$$\delta = \min \{ \delta_i : 1 \leqslant i \leqslant n\} > 0.$$
And thus as a consequence of the finiteness of $\mathcal{C}$ and the condition that any two [distinct] members of $\mathcal{C}$ have a positive distance we obtain that there is a $\delta > 0$ such that
$$\operatorname{dist}\biggl(A, \operatorname{Cl}\Bigl(\bigcup \bigl(\mathcal{C}\setminus \{A\}\bigr)\Bigr)\biggr) \geqslant \delta \tag{$\ast$}$$
for every $A \in \mathcal{C}$. This of course implies $A \cap \operatorname{Cl}\Bigr(\bigcup \bigl(\mathcal{C}\setminus \{A\}\bigr)\Bigr) = \varnothing$, but it is much stronger.

The book further goes on to add that for uniform continuity, the condition $\operatorname{dist} (A,B) > 0$ cannot be weakened to $\bar A \bigcap \bar B = \varnothing$

Right, there can be disjoint closed sets in $X$ whose distance is $0$. For example, in $\mathbb{R}^2$ (with the Euclidean metric) consider $A = \{(x,0) : x \in \mathbb{R}\}$ and $B = \{(x,y) : x\cdot y = 1\}$. Let $f\lvert_A \equiv 0$ and $f\lvert_B \equiv 1$. Then $f$ is continuous but not uniformly continuous on $A \cup B$.
A: Note in the condition for uniform continuity $\mathcal C$ is a finite collection of sets, while in the ordinary continuity condition $\mathcal C$ is not required to be finite. 
Because of that finiteness, the uniform continuity condition implies the ordinary continuity condition.
