Show that the function $f(x)=\frac{1}{x}$is uniformly continuous on the set $A:=[a,\infty )$ where a is positive constant please check my proof 
Let $\epsilon >0 \delta >0$ and a is arbitary positive constant in interval
$$|x-a|<\delta \leftrightarrow |f(x)-f(a)|<\epsilon $$
$$\leftrightarrow |\frac{1}{x}-\frac{1}{a}|<\epsilon $$
$$\leftrightarrow |\frac{a-x}{xa}|<\epsilon $$
$$\leftrightarrow \frac{1}{a}|\frac{a-x}{x}|<\epsilon $$
$$|\frac{a-x}{x}|<a\epsilon $$
choose $\delta =a\epsilon $
then 
$|\frac{1}{x}-\frac{1}{a}|<\delta \leftrightarrow |\frac{1}{x}-\frac{1}{a}<a\frac{\epsilon }{a}=\epsilon $
then it is uniformly conitinouos 

As pointed out in a comment, you're only looking at values near $x=a$. If you prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x)-f(a)| \le \epsilon$ whenever $|x-a| \le \delta$ (which you did not, your proof is confused), then this would only imply the continuity of $f$ at $a$, not more. 
I suggest another approach: note that $f$ is continuous in $[a,+\infty)$ (if $a>0$) and $\lim_{x \rightarrow +\infty} f(x) = 0 $. These two properties combined together give the uniform continuity of $f$ in $[a,+\infty)$. The main idea is to combine Heine-Cantor theorem (a continuous function on a compact set is uniformly continuous) with the vanishing condition. Let me know if you manage to conclude or you need a complete proof.
