$\epsilon$-$\delta$ proof that $\lim_{x \to 1} \sqrt{x} = 1$

I'm trying to teach myself how to do $\epsilon$-$\delta$ proofs and would like to know if I solved this proof correctly. The answer given (Spivak, but in the solutions book) was very different.

Exercise: Prove $\lim_{x \to 1} \sqrt{x} = 1$ using $\epsilon$-$\delta$.

My Proof:

We have that $0 < |x-1| < \delta $.

Also, $|x - 1| = \bigl|(\sqrt{x}-1)(\sqrt{x}+1)\bigr| = |\sqrt{x}-1||\sqrt{x}+1| < \delta$.

$\therefore |\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|}$

Now we let $\delta = 1$. Then \begin{array}{l} -1<x-1<1 \\ \therefore 0 < x < 2 \\ \therefore 1 < \sqrt{x} + 1<\sqrt{2} + 1 \\ \therefore \frac{1}{\sqrt{x} + 1}<1. \end{array}

We had that $$|\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|} \therefore |\sqrt{x}-1|<\delta$$

By letting $\delta=\min(1, \epsilon)$, we get that $|\sqrt{x}-1|<\epsilon$ if $0 < |x-1| < \delta $.

Thus, $\lim_{x \to 1} \sqrt{x} = 1$.

Is my proof correct? Is there a better way to do it (still using $\epsilon-\delta$)?

The proof is correct but can be simplified. You don't need the part "Now let $\delta=1$...". In fact it is always true that $$ \frac{1}{\sqrt x + 1} \le 1 $$ since $\sqrt x \ge 0$.

Also, a matter of style. In the first line you don't have $0 < |x-1|<\delta$ but you suppose it (this is because $\delta$ is not already been given, but has to be found yet). The same when you write "let $\delta = 1$" you should write "if $\delta \le 1$ ..."

Your proof is correct.
We can also adopt the following:
Since $|\sqrt x-1|\lt \epsilon$ is equivalent with $1-2\epsilon+\epsilon^2\lt x\lt 1+2\epsilon+\epsilon^2$, we can choose $\delta$ so that $0\lt\delta\lt \min\{|-2\epsilon+\epsilon^2|,|2\epsilon+\epsilon^2|\}$.
Hope this helps.