Uniform continuity of continuous functions on compact sets Assume that $f: \mathbb R \rightarrow \mathbb R$ is continuous function on the compact set $A$. 
Does for any $\varepsilon >0$ exist a $\delta >0$, such that 
$$
\lvert\, f(x)-f(y)\rvert<\varepsilon \,\,\,\,\,\,\textrm{for every}\,\,\,\, x,y\in A,\,\,
\text{with}\,\,\, 
\lvert x-y\rvert<\delta?
$$

I think this is called Heine's theorem, here goes my attempt:
First note that your condition is equivalent to: $$\exists\delta>0 : \forall\varepsilon>0, x,y\in A  \Rightarrow|f(x)-f(y)|<\varepsilon, \text{if } |x-y|<\delta$$
Which also means $$\forall\varepsilon>0, (x_n),(y_n)\subset A, \ \lim(x_n-y_n)=0 \Rightarrow |f(x_n)-f(y_n)|<\varepsilon$$
Also I'll use the definition of compact by sub-successions, which is that: $$\forall (x_n)\subset A, (x_n) \text{ succession, then there exists a convergent sub-succession} (x_{n_k})\subset A, \text{and} \\ \lim x_{n_k}\in A$$
So, first suppose that $f$ is not uniformly continuos, so we negate the previous statement:
$$
\exists\varepsilon>0, (x_n),(y_n)\subset A : \ \lim(x_n-y_n)=0 \Rightarrow |f(x_n)-f(y_n)|\ge\varepsilon \tag{#}
$$
So, let $(x_{n_k}), (y_{n_k})$ be convergent sub-successions, with $(x_{n_k})\to x_0$. Since $\lim(x_n-y_n)=0 \Rightarrow (y_{n_k})\to x_0$. Now use the fact that $f$ is continuos$^*$, in which case $$(f(x_{n_k}))\to f(x_0), f(y_{n_k})\to f(x_0). \text{Therefore, }\lim(f(x_{n_k})-f(y_{n_k}))=0.$$
Using the definition of limit this contradicts condition $\text{(#)}$, therefore $f$ is uniformly continuos. Note that by replacing "$|$" with "$||$" we obtain the generalization for metric spaces! (because the succession characterizations also hold).
$*\text{using $f$ continuous} \Leftrightarrow \forall (x_n)\mid\lim(x_n)=p, \ \lim f(x_n)=f(p). $
