Clarification of a proof that $x^2$ is continuous everywhere. So I've been looking at the proof for this question and was wondering if some one could help me walk through the reasoning or could look through my proof.

We fix $\epsilon >0$, and observe that $|x^2 -c^2|= |x-c||x+c|$. Suppose that $\delta < 1$ so that $|x-c| < \delta < 1$.

(First I want to discuss this assumption - my understanding is that we can make this assumption because if in reality $\delta \geq 1$, a smaller $\delta$ will always 'fit'. I.e. then, $|x-c| < 1 \Rightarrow |x-c| < \delta$, so this assumption is safe. Is this reasonable?)

Next, we observe $|x+c| = |x-c+2c| \leq |x-c| + 2|c| < 1 + 2|c|$ 
  Thus using difference of two squares I can say that:
  $|x^2-c^2| < \epsilon$
  and that
  $|x^2-c^2| = |x-c||x+c| < \delta (1+2|c|)$ 

Here I want to ask, what is the relation between $\epsilon$ and $\delta (1+2|c|)$, and why. I've seen answers stating that $\epsilon$ is greater but I can't why this is the case.

From here they generally conclude that we take $\delta$ to be the smaller of $\frac{\epsilon}{1+2|c|}$ and $1$. 

But if the latter, doesn't this feel contradictory given our assumtion that $\delta < 1$ Or are we simply saying that such a value of $\delta$ is 'sufficient'. From my understanding our logic here is that we need to satisfy both conditions we placed on $\delta$ thus we set it to be the smallest?
 A: The key is really that the first part of the proof really isn't the proof. It's just giving a way to find the proof.
The actual proof is:

Given $\epsilon>0$, let $\delta=\min\left(\frac{\epsilon}{1+2|c|},1\right).$
  Then if $|x-c|<\delta$ we have:
  $$ |x-c|<\delta\leq \frac{\epsilon}{1+2|c|}\\
|x+c|\leq |x-c| + 2|c|<\delta+2|c|\leq 1+2|c|$$
  So:
  $$|x^2-c^2|=|x-c||x+c|< \frac{\epsilon}{1+2|c|} \cdot \left(1+2|c|\right)=\epsilon$$

There is a tendency to write these proofs in reverse, to explicate how you got the component conditions for $\delta$. But that is not necessary[*], it is just exposition for learning purposes, not a natural part of the proof.
As you can see, we are using both conditions, and it is quite clear what both conditions are doing. We are trying to bound a product by bounding the two factors. The second line is not bounded if you don't put an additional bound on $\delta$.
[*] Not necessary as part of a formal proof, but it might be preferred on homework, because it is like "showing your work" in a computation.
By the way, you could rewrite the proof as:

Given $\epsilon>0$, let $\delta=\min\left(\frac{\epsilon}{\pi+2|c|},\pi\right).$
  Then if $|x-c|<\delta$ we have:
  $$ |x-c|<\delta\leq \frac{\epsilon}{\pi+2|c|}\\
|x+c|\leq |x-c| + 2|c|<\delta+2|c|\leq \pi+2|c|$$
  So:
  $$|x^2-c^2|=|x-c||x+c|< \frac{\epsilon}{\pi+2|c|} \cdot \left(\pi+2|c|\right)=\epsilon$$

The number $1$ is a red herring. You could have picked any positive value initially to bound $|x+c|$, as long as you change the second bound, as well. The number $1$ is just the easiest positive number to write.
There is no one "right" $\delta$.
A: Without saying anything else, you first assume $\delta<1$. You need not to say anything about this than just saying this is an assumption. After all, you are required to find some $\delta$ when you are given some $\varepsilon>0$.
Now assuming $\delta<1$, you found that
$$
|x^2-c^2|<\delta (1+2|c|)
$$
Now since $c$ is fixed, $(1+2|c|)$ is a fixed constant, and since we are given $\varepsilon>0$, then clearly $\displaystyle \frac{\varepsilon}{1+2|c|}$ is a fixed postive constant. 
Therefore, at least one of 
$$
1>\frac{\varepsilon}{1+2|c|} \;\;\; \text{or} \;\;\; 1\le \frac{\varepsilon}{1+2|c|}
$$
happens, so the quantity $\delta=\min\{1,\frac{\varepsilon}{1+2|c|}\}$ is well defined. Then naturally $\delta\le 1$, and so as you've shown in this case
$$
|x^2-c^2|\le\delta (1+2|c|) \le \frac{\varepsilon}{1+2|c|} (1+2|c|) = \varepsilon
$$
