If f is unformly continuous on two sets, show that f is also uniformly continuous on the union of two given sets

The definition says $\forall\varepsilon>0\ \exists\delta>0\cdots\cdots\cdots\cdots$. To prove that a function is uniformly continuous, you need to find $\delta$ as a function of $\varepsilon$ and prove that it's small enough. You know you've got $\delta_1$ that's small enough on one set and $\delta_2$ that's small enough on the other. Which one is smaller might depend on $\varepsilon$. However $\min\{\delta_1,\delta_2\}$ will be small enough on both sets.

Later note, per comments: Let's make $\delta_A$ small enough so that if $x,y\in A$ and $|x-y|<\delta_A$ then $|f(x)-f(y)|<\varepsilon/2$, and if $x,y\in B$ and $|x-y|<\delta_B$ then $|f(x)-f(y)|<\varepsilon/2$.

Let $\delta=\min\{\delta_A,\delta_B\}$.

If $x,y\text{ both}\in A$ or $\text{both}\in B$, and $|x-y|<\delta$ that does it, as above.

If $x\in A$ and $y\in B$, then the distances from $x$ to the boundary point $b$, and from $y$ to $b$, are less than $\delta$, so $$|f(x)-f(y)| \le|f(x)-f(b)|+|f(b)-f(y)|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$

So in all cases, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$.