Why two Inequalities are true $A$ is a subset of real numbers.

Consider the set $A$ of all real
  numbers  $x$  such that $x^2 \le 2$.

This set is nonempty and bounded from above, for example by 2. Call the $\sup(A),\; y$. Then $y^2 = 2$, because the other possibilities 
$y^2 \lt 2$  and  $y^2 \gt 2$ both lead to a contradiction. 

Assume that $y^2 \gt 2$, so there exists $z \gt 0$ such that $y^2 − 4z \gt 2$.
    Then:

$$ (y − z)^2 = y^2 − 2zy + z^2 > y^2 − 4z > 2 \tag{1} $$

Therefore $y − z$ is also an upper bound of $A$ and $y$ cannot be the least upper bound.
If $y^2 \lt 2$, so there exists $c \gt 0$ such that $y^2 + 5c \lt 2$ and $c^2 \lt c$. 
  Then:

$$ (y + c)^2 = y^2 + 2cy + c^2 < y^2 + 5c < 2. \tag{2} $$

Therefore also $y + c \in A$ and $y$ cannot be the least upper bound.

I apologize for this basic question.  The inequalities labelled 1 and 2 appear to be true to me, but I can't see why they are true logically. I don't see the "then" logical link.
 A: You can think (1) in this way: Let $z$ be close enough to $(y^2-2)/4$, it is convenient to think they are equal. Then you will find (1) is right, of course, take $z$ smaller is ok!
A: In (1), the equality part is easy and the second inequality is just an assumption. So, you only need to prove that $y^2-2zy+z^2 > y^2-4z$ which is equivalent to $z^2-2zy+4z > 0$ and, since $z>0$, this is equivalent to $z-2y+4 > 0$, or, in other words, $z > 2y-4$. So, you only need to prove $z > 2y-4$. Now, since $y$ is upper-bounded by $2$, we have $y < 2$ which is equivalent to $2y - 4 < 0$ and since $z > 0$, we have $z > 2y-4$ as required.
Similarly for (2), inequality $y^2+2cy+c^2 < y^2+5c$ is equivalent to $c^2 + 2cy - 5c < 0$ which, since $c > 0$ is equivalent to $c + 2y - 5 < 0$. This holds because $c < 1$ (since $c^2 < c$ and $c > 0$) and $y < 2$ ($y$ is upper-bounded by $2$). Therefore, $c + 2y < 1 + 2 * 2 = 5$.  
A: The expressions (1) and (2) might be easier to follow if they are expanded to
$$\begin{align}
(y-z)^2&=y^2-2yz+z^2\\
&=y^2-4z+4z-2zy+z^2\\
&=y^2-4z+z(z+2(2-y))\\
&\ge y^2-4z\\
&\gt2
\end{align}$$
and
$$\begin{align}
(y+c)^2&=y^2+2cy+c^2\\
&\lt y^2+2cy+c\\
&=y^2+5c-5c+2cy+c\\
&=y^2+5c-2c(2-y)\\
&\le y^2+5c\\
&\lt2
\end{align}$$
Note, both expansions use the fact that $y\le2$ (i.e., $2$ is an upper bound for the set $A$); the second makes explicit use of the condition $c^2\lt c$. The key idea is the insertion of $0$, in the form $-4z+4z$ and $5c-5c$, respectively.
