Why is delta assumed to be 1 in proving that $x^2$ is continuous? 
The function $f(x)=x^2$ is continuous everywhere.

I wrote all this for good measure, but my question is that in most proofs the calculations below are avoided and instead $\delta$ is chosen to be less than $1$ or chosen to be the $\mathrm{min}(1,\frac{\varepsilon}{2|a|+1})$. For example, as done in this question Show Continuity Using Epsilon Delta Definiton .
How do we know to choose $\delta < 1$ or $\mathrm{min}(1,\frac{\varepsilon}{2|a|+1})$ as done in the answer of the linked question? Shouldn't the calculations below be necessary in any conclusive proof of the question.
Using the definition of continuity as
$$\forall \epsilon >0\  \forall x_0 >0 \in S \  \exists \delta > 0\  \forall x \in S [\ | x-x_0| < \delta \Rightarrow \ |f(x)-f(x_0)| <\epsilon ]$$, I realize that we have to define our $\delta$ both in terms of $x_0$ and $\epsilon$. So $$| x-x_0| < \delta$$ $$ -\delta +x_0 < x  < \delta+x_0$$  $$ -\delta +2x_0 < x + x_0  < \delta+ 2x_0$$
$$-\delta +2|x_0| \le-\delta +2x_0 < x + x_0  < \delta+ 2x_0 \le \delta+ |2x|_0$$
$$-\delta +2|x_0|  < x + x_0  < \delta+ |2x_0|$$
$$ |x + x_0|  < \delta+ |2x_0|$$
Now since $| x-x_0| < \delta$, it follows  $$| x-x_0||x + x_0| < \delta[\delta+ |2x_0|]$$ 
$$|f(x)-f(x_0)| < \delta^{2}+ 2|x_0|\delta \ .$$
So we let $$\delta^{2}+ 2|x_0|\delta =\epsilon$$
$$\delta^{2}+ 2|x_0|\delta - \epsilon=0$$ and solve using the quadratic formula, that is, 
$$\delta=  \frac{ (-2|x_0|)\pm \sqrt{ {(2|x_0|)^2}+4\epsilon}}{2} = \frac{-|x_0|\pm \sqrt{ (|x_0|^2)+ \epsilon}}{1}.$$
Since $\delta > 0$ we let $\delta = -|x_0| $ + $  \sqrt{ (|x_0|^2)+ \epsilon}$.
 A: The point of limiting $\delta$ to be less than $1$ is that the further out from $x_0$ you go, the steeper the graph gets. Limiting $\delta$ means you don't have to deal with what happens far away from the point where you're checking for continuity.
If you don't limit $\delta$, as you've experienced, the dependence of $\delta$ on $\epsilon$ becomes convoluted precisely to take into account the cases where $\epsilon$ happens to be large. Deciding that "no matter what happens, I will never make my $\delta$ larger than $1$" before you even do any calculations lets you focus on the cases that actually matter: the small $\epsilon$'s, where a linear dependence between $\epsilon$ and $\delta$ usually is good enough.
Note that there is nothing special about $1$ in this context. You are free to choose any positive number. Other functions, however (like $\frac1x$), might require a more specific initial bound on $\delta$, possibly even one that depends on $x_0$.
A: I always feel it is strongly advisable to work all the inequalities out as "rough working" before you attempt to write the formal proof.  To show that
$$\lim_{x\to a}x^2=a^2$$
you could do the following:
$$\eqalign{|x^2-a^2|
  &=|x-a|\,|x+a|\cr
  &=|x-a|\,|(x-a)+2a|\cr
  &\le|x-a|\bigl(|x-a|+2|a|\bigr)\cr}$$
- the reason for these steps is that we want to write the expression in terms of $|x-a|$.  Now IMHO the crucial point is that

$|x-a|<\delta$, and $\delta$ is whatever we want it to be (as long as it makes the proof work).

So $(*)$ I am going to assume $|x-a|<1$, and my inequality implies
$$|x^2-a^2|\le|x-a|(1+2|a|)\ .$$
My aim is that the RHS should be less than $\varepsilon$, so I am also going to assume
$$|x-a|<\frac{\varepsilon}{1+2|a|}\ .$$
Therefore my choice for $\delta$ will be
$$\delta=\min\Bigl(1,\,\frac{\varepsilon}{1+2|a|}\Bigr)\ ,$$
because this guarantees that both the inequalities I need will be true.
Notice, however, that there is really nothing special about the choice $|x-a|<1$.  Starting again from $(*)$, I could have assumed $|x-a|<12345$; then through similar working (please try it for yourself) I would have ended up choosing
$$\delta=\min\Bigl(12345,\,\frac{\varepsilon}{12345+2|a|}\Bigr)\ .$$
I could even have assumed $|x-a|<\varepsilon$, leading to
$$\delta
  =\min\Bigl(\varepsilon,\,\frac{\varepsilon}{\varepsilon+2|a|}\Bigr)\ .$$
Exercise.  Here's one which looks even simpler, but it doesn't work - can you explain why?  Assume $|x-a|<|a|$ and hence choose
$$\delta=\min\Bigl(|a|,\,\frac{\varepsilon}{3|a|}\Bigr)\ .$$
Note that the above will be rough working, and the actual proof will be as follows.

Let $\varepsilon>0$.
  Choose $\delta=\min\Bigl(1,\,\dfrac{\varepsilon}{1+2|a|}\Bigr)$; clearly $\delta>0$.
  Suppose that $0<|x-a|<\delta$.  Then
  $$|x-a|<1\quad\hbox{and}\quad |x-a|<\frac{\varepsilon}{1+2|a|}\ ,$$
  and therefore
  $$|x^2-a^2|=\cdots<\varepsilon\ .$$

I'll leave you to fill in the dots.  Hope this helps.
A: This is not exactly an answer, but it is too long for a comment. 

The definition of continuity which you have written is correct, but at the same time extremely difficult for any beginner to understand and even much less useful in application. A person who can really understand such definition full of logical symbols like $\forall, \exists$ is already too experienced in the art of calculus to bother to prove the continuity of $f(x) = x^{2}$.
Also note that the definitions of limit/continuity are based on the idea of inequalities, but these definitions should not be considered as a smart manipulation of algebraic inequalities. Rather these definitions build on the idea of "order" and "density" of the real numbers.
Any approach involving algebraic manipulation of inequalities will lead you nowhere. Thus to prove the continuity of $x^{2}$ at $a$ is not equivalent to solving the in-equation $$|x^{2}-a^{2}|<\epsilon$$ to get the solution in the form $$|x-a|<\delta$$ By the definition if one $\delta$ works any smaller $\delta$ also works. Hence solving for $\delta$ in terms of $a, \epsilon$ does not work. It is more important to show that a $\delta$ exists and if desired we can find an expression for $\delta$ in terms of $a, \epsilon$. Also since $\delta$ is not unique we can have multiple and equally correct expressions for $\delta$. Thus taking an expression from David's answer I can say that instead of choosing $$\delta=\min\left(1,\frac{\epsilon}{1+2|a|}\right)$$ we can choose any positive number $\delta$ which is less than the expression on the right in the above equation. 

Next I point another issue in your approach which many people may miss (this was first pointed out by Spivak in his calculus book). You have obtained an expression for $\delta$ using square roots. But note that existence of square roots is dependent on the fact that $x^{2}$ is continuous. Hence any use of square root operation is circular here. In general, to prove the continuity of a function we cannot use the inverse of that function. 
