Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element in $\mathbb{Q}$ 
Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element in $\mathbb{Q}.$

I assume that $y=\inf A=$ least element. Then $y\in A$ or $y$ does not belongs to A. 
Case I: Let $y\in A$; 
Considering $k=\frac{y^2-2}{2y}$, I have shown $y-k \in \mathbb{Q}^+$ (as $y-k>0$ and rational). Also $(y-k)^2>2$ implying $y-k\in A$. But $y-k<y$ and $y=\inf A$, a contradiction. So $A$ has no least element in this case.
Case II: Let $y$ that does not belongs to $A$. Then either $y^2=2$ or $y^2<2$.
My questions:


*

*Please help me to complete case II. 

*Is there any other way to solve it?

 A: Here is a quick answer on request of OP which is just an explanation of my comment. Let $x\in A$ then $x^{2}>2$ and now consider the rational number $y=(3x+4)/(2x+3)$. First note that $$x-y=\frac{2(x^{2}-2)}{2x+3}>0$$ so that $y<x$. Next I prove that $y\in A$. We can see that $$y^{2}-2 = \frac{(3x+4)^{2}-2(2x+3)^{2}}{(2x+3)^{2}}=\frac{x^{2}-2}{(2x+3)^{2}}>0$$ so that $y^{2}>2$. We thus see that if $x\in A$ then there is a smaller $y$ such that $y\in A$ and therefore $A$ has no least member.
It is best not to give a proof using the concept of infimum/supremum because these may not exist for all sets in $\mathbb{Q} $. The problem is just a specific property of set $A$ of rational numbers which is easily proven via simple algebra. 
A: I think the real source of confusion here is the poorly worded question:

(1) Ambiguous: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element in $\mathbb{Q}.$

A least element is a property of an ordered set $X$. No other data is involved. There's no such relation as "least element of $X$ in $Y$" for some other set $Y$. So a much more sensible question would be simply:

(2) Good: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element.

That would make it clear that you only have to consider Case I. In my opinion, it is reasonable to interpret question (1) as meaning (2). This is assumed by the other answers on this page.
Or we could ask a different question:

(3) Also good, and harder: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no infimum in $\mathbb{Q}.$

It seems like the wording of question (1) made you assume that the author meant to ask (3). Perhaps they did! To answer (3), you do need to consider both Case I and Case II. Some of the other answers on this page address how you would do that.
As a final note, if we want to be charitable to the author, we could interpret (1) more straightforwardly to mean this:

(4) Silly: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element $y$ such that $y\in\mathbb{Q}.$

This is silly because $A$ is defined as a subset of $\mathbb{Q}$, so the question immediately reduces to (2). No one would actually write (4). But then, they shouldn't have written (1) in the first place.
A: [Edited to engage directly with "Please help me to complete Case II"]
You write, "I assume that $y=\inf A=$ least element." The difficulty with this has been noted in the comments, but since it is still in the body of the question, I want to address it: 
The infimum of a set is not the same thing as the least element of a set. The infimum of the set $P$ of positive reals, for example, is zero, but the least element of $P$ is ... well, it isn't, it doesn't exist; $P$ has no least element.  
What is true is that if a set (of reals) has both a least element and an infimum then they are equal. 
Now, $A$ is defined to be a subset of the rationals. It follows that if $A$ has a least element, then that least element is rational, so we are just trying to prove that $A$ has no least element. 
Now in Case II, you are assuming $y$ is not in $A$. If by $y$ you mean $\inf A$, then you are assuming $\inf A$ is not in $A$, which is the same as saying that $A$ has no least element, and we're done. If by $y$ you mean the least element of $A$, then Case II doesn't exist, since it makes no sense to ask what happens if the least element of $A$ isn't an element of $A$. So, no matter what you mean by $y$, there is nothing needing doing in order to complete Case II. 
[Previous version of answer:]
You have already shown that if $y$ is any element of $A$ then there is a smaller element of $A$ (namely, $y-k$). This proves there is no smallest element of $A$. 
A: Your reasoning in Case I is actually all you need here! To show a set doesn't have a least element, you just have to show that any given element (call it $y$) isn't the least element--that is, that there exists some $x$ in the set with $x < y$.
I'll add a more concise/straightforward proof below, but first I'll work with what you have so far (answering "Please help me to complete case II.").
Let $y = \inf A$ and suppose $y \notin A$ (Case II). Let $x$ be an arbitrary element of $A$; then by the definition of $\inf$, there exists some $z \in A$ with $y \leq z < x$. Then $x$ is not the least element of $A$. Since $x$ was arbitrary, this means there is no least element of $A$, which is what we want to show.
As for your other request ("Is there any other way to solve it?"), you can do this (1) without using $\inf$ and (2) more concisely/directly:
Proof: Let $y \in A$, and let $k = \frac{y^2-2}{2y}$. Since $y \in \mathbb Q$ and $y^2 > 2$, $k \in \mathbb Q^+$. You can show* $(y-k)^2 > 2$, so $y-k \in A$, which means $y$ is not the least element of $A$. Since $y$ was an arbitrary element of $A$, this means $A$ has no least element. $\blacksquare$
*You used this in your Case I, but I wanted to verify for myself: 
\begin{align*}
(y-k)^2 &= \left(\frac{2y^2}{2y} - \frac{y^2-2}{2y}\right)^2 = \frac{((y^2-2)+4)^2}{4y^2} \\
& = \frac{(y^2-2)^2 + 8(y^2-2) + 16}{4y^2} \\
& = \frac{(y^2-2)^2}{4y^2} + \frac{8y^2}{4y^2} > 2
\end{align*}
