I've tried to proof that $f(x) = x^2-2$ is continuous for all $x \in \mathbb{R}$ using the $\delta -\epsilon$ definiton of continuity, I think my proof is correct. But it's not the same as my textbook's. Would someone be willing to check whether this proof is correct? Here's my proof: Let $c \in \mathbb{R}$ be randomly given $f$ is continuous in $c$ if $\forall \epsilon \in \mathbb{R}_{>0} \exists \delta \in \mathbb{R}_{>0}$ so that $\forall x \in \mathbb{R}$ it is the case that $|x - c| < \delta \implies |f(x)-f(c)| < \epsilon$ such a $\delta$ exists, choose namely $\delta = \frac{\epsilon}{2\cdot|x+c|}$. This works, because $|f(x)-f(c)| = |(x^2-2)-(c^2-2)| = |x^2-c^2| = |(x-c)(x+c)| = |x-c| \cdot |x+c|$ But remember that $|x-c| < \frac{\epsilon}{2\cdot|x+c|}$, so $|x-c| \cdot |x+c| \leq \frac{\epsilon}{2\cdot|x+c|} \cdot |x+c|= \frac{\epsilon}{2} <\epsilon$
So it is true that$\forall \epsilon \in \mathbb{R}_{>0} \exists \delta \in \mathbb{R}_{>0}$ $|x - c| < \delta \implies |f(x)-f(c)| < \epsilon$, and considering $c \in \mathbb{R}$ was randomly given, this stament holds $\forall c\in \mathbb{R}$, so f is continuous on the whole of $\mathbb{R}$.
As pointed out, you cannot have your $\delta$ depend on $x$. Instead, with these types of proofs, it's easier to start with what you want. delta-epsilon proofs tend to be easier for me when I start by working backward. We can get around the problem with your original proof with the following trick:
WlOG assume that $c>0$.
Goal: $|x^{2}-2-c^{2}+2|<\epsilon$
$|(x+c)(x-c)|<\epsilon$.
$|(x+c)|\cdot|x-c|<\epsilon$.
Suppose that $\delta<1$.
Then $|x-c|<1 \implies 2c-1<x+c<2c+1$
then $|x-c||x+c|<\epsilon \implies \delta\cdot(2c+1)<\epsilon $
...Can you see where this is headed? The idea is to get rid of the $x$ by only dealing with the entire $|x+c|$ term, instead of using $x$ explicitly. By assuming a pretty small delta (somewhat arbitrary), you can use this fact to get a neat expression for $\delta$. Notice my "WLOG", otherwise $|x+c|<2c+1$ isn't justified. Although you could very easily take the max of the two values.