How to explain this contradiction with infimum and $\epsilon$? If we have $\inf A$ of a set $A$, then $\inf A+ϵ$ is not a lower bound of $A$. There exists some elements $x$ of $A$ such that $x\in A$ and $x\in[\inf A,\inf A+\epsilon)$. However, $x<\inf A+\epsilon$ implies $x\leq\inf A$ which means these $x$ are lower bound of $A$. It contradicts the definition of infimum. How to explain it?   
My Attempt: Suppose the only element in $[\inf A,\inf A+\epsilon)$ is the $\inf A$, then it is true that $\inf A \leq \inf A$ and $\inf A\in A$. However, it is not always true that $\inf A \in A$. If $\inf A\notin A$, then I have a contradiction that $x\in A$ and $x \leq \inf A$ at the same time.
 A: I suspect you learned the theorem "If $x<y+\epsilon$ for all positive $\epsilon$ then $x\leq y$" and have confused it with the falsehood "for all positive $\epsilon$, if $x<y+\epsilon$ then $x\leq y$."
A: You seem to be assuming something very special about $\epsilon$ when you write

$x<inf A+\epsilon$ implies $x\le inf A$.

This would only be true if $\epsilon$ were "as small as possible" so that there would be no room between $inf A$ and $inf A + \epsilon$ for any more elements of $A$. But there is no such number: for any positive $\epsilon$, the number ${\epsilon\over 2}$ is also positive but $<\epsilon$.
For a concrete example, take $A$ to be the set of positive reals. Then $inf A=0$, and for every positive $\epsilon$ the number $\epsilon\over 2$ is an element of $A$ which is less than $inf A+\epsilon$.

As a far more minor aside, you're also implicitly assuming that $A$ does not have a least element - if $inf A\in A$ then for any positive $\epsilon$ the number $inf A$ is an element of $A$ less than $inf A+\epsilon$, so even if there were an $\epsilon$ with the above property - which, again,there isn't - we wouldn't have a contradiction in this case.
A: If $x < K+\epsilon$ for ALL $\epsilon > 0$ then $x \le K$.
But if $x < K + \epsilon$ for SOME $\epsilon > 0$ then $K < x  < K + \epsilon$ is very much possible.  (For example if $x = K + \frac \epsilon 2$)
It is true that for ALL $\epsilon > 0$ then $\inf A + \epsilon $ is not a lower bound and therefore that if $w$ is a lower bound of $A$ then $w \le \inf A$.  But for every $\epsilon > 0$ there will be some $x_\epsilon$ so that $\inf A \le x_\epsilon < \inf A + \epsilon$.  That is true for THAT $\epsilon > 0$.  
However there are not any $x$ so that $\inf A < x < \inf A + \epsilon$ for ALL $\epsilon$.  But that is not what the statement is saying.  The statement is saying for THAT $\epsilon$ there is an $x$ so that $\inf A \le x < \inf A + \epsilon$.  Having $\inf A < x$ for THAT $\epsilon$ will not be a contradiction.
A: The difference between all $\epsilon$ and some $\epsilon$ needs to be made explicit or abundantly clear (if needed) in your language.
You can write:

For every $\epsilon>0$ the number $\inf A+\epsilon$ is not a lower bound for $A$ and hence there exists a member $x_\epsilon\in A$ corresponding to $\epsilon$ such that $$\inf A\leq x_\epsilon <\inf A+\epsilon$$ 

The above statement is true and you don't get the conclusion $x_\epsilon\leq \inf A$ because the number $x_\epsilon$ is not fixed but rather depends on the value of $\epsilon$.
The logical part of definitions in elementary analysis is not difficult and best handled by writing in natural language rather than using logical symbols and technical jargon. Reserve these for those occasions when you want to intimidate your readers.
Also if you are unlucky to have read a theorem like (which is most likely the source of your confusion here) 

(Unnecessary) Theorem: If $x, y$ are real numbers such that $x<y+\epsilon$ for all $\epsilon>0$ then $x\leq y$.

then do translate it into

(Obvious) Fact: There is no least positive real number.

Also treat the above obvious fact on par with there is no least positive rational number and there is no greatest positive integer. 
