Let $(X,d)$ be a metric space. Let $K \subset X$ compact and $A \subset X$: $\exists \delta>0$ such that $d(a,b)>\delta$ for all $a,b \in A$ with $a \neq b$. Consider in $K \cup A$ the induced metric BY $(X,d)$. Prove that every continuous function $f: K \cup A \to \mathbb R$ is uniformly continuous.
This is what I've tried:
The problem can be divided in two cases:
$1)$ If $K \cap A=\emptyset$, then $d(K,A)=r>0$. Let $f:K \cup A \to \mathbb R$ a continuous function and let $\epsilon>0$. Since $f$ is continuous and $K$ is compact, then $f|_K$ is uniformly continuous, which means there is $\delta_0$ such that $d(x,y)<\delta_0 \implies |f(x)-f(y)|<\epsilon$ for all $x,y \in K$.
On the other hand, if $d(x,y)<\dfrac{\delta}{2} \implies x=y \implies |f(x)-f(y)|=0<\epsilon \space \forall x,y \space \in A$ Consider $\delta'=\min\{\delta_0,\dfrac{\delta}{2},\dfrac{r}{2}\}$. Then, since $d(x,y)<\delta' \implies x,y \in K$ or $x,y \in A$, by what we've seen a few lines above, we have that $d(x,y)<\delta' \implies |f(x)-f(y)|<\epsilon$ for all $x,y \in K \cup A$. From here it follows that every continuous function $f: K \cup A \to \mathbb R$ is uniformly continuous.
For case $2)$ I have some doubts:
Case $2)$ is the case $K \cap A \neq \emptyset$. I am just going to skip the case $A \subset K$ because this is a trivial case. If I express $K \cup A$ as $K \cup (A \setminus (K \cap A))$, wouldn't $A \setminus (K \cap A)$ be still discrete? If this is true, wouldn't case $2)$ be reduced to case $1)$?
I would appreciate if someone could tell me if my approach is correct and if someone has a simpler, shorter or different idea on how to prove the statement, I welcome him/her to tell me.
Suppose not. Then there exists a $\varepsilon>0$, such that for every $n\in\mathbb N$, there exist $x_n,y_n\in K\cup A$, with $d(x_n,y_n)<1/n$ and $|f(x_n)-f(y_n)|\ge\varepsilon$.
If $K$ contains infinitely many terms of the sequence $\{x_n\}_{n\in\mathbb N}$, then compactness of $K$ guarantees the existence of a convergent subsequence $x_{n_k}\to x\in K$, and since $d(x_{n_k},y_{n_k})\to 0$, then $y_{n_k}\to x$ as well, and because of continuity of $f$ $$ f\big(x_{n_k}\big)\to f(x) \quad \text{and}\quad f\big(y_{n_k}\big)\to f(x), $$ which contradicts the fact that $|f(x_n)-f(y_n)|\ge\varepsilon$. We also arrive to a contradiction if we assume that If $K$ contains infinitely many terms of the sequence $\{y_n\}_{n\in\mathbb N}$.
If on the other hand $K$ contains only finitely many terms of each of the sequences $\{x_n\}_{n\in\mathbb N}$ and $\{y_n\}_{n\in\mathbb N}$, then there exists an $n_0\in\mathbb N$, such that $$ n\ge n_0\quad\Longrightarrow\quad x_n,y_n \in A. $$ But since $d(x_n,y_n)<1/n$, then for sufficiently large $n$, we have $\delta>1/n$, which would imply that $x_n=y_n$ and hence $f(x_n)=f(y_n)$, contradicting the fact that $|f(x_n)-f(y_n)|\ge\varepsilon$. $\Box$