<s>[INST] Proof of uniform continuity I seem to have hit a dead-end in the following proof.

Define $f:\mathbb{R}\to\mathbb{R}$ by:
$f(x)=\frac{1}{1+x^2}$
Show that $f$ is uniformly continuous.

My proof:

Let $x_{0}\in \mathbb{R}$.
Also let $\epsilon >0$
Choose $\delta = ?$
Then, for $x\in \mathbb{R}$, such that $|x-x_{0}|<\delta$, we have:
|$f(x)-f(x_{0})|=|\frac{1}{1+x^2}-\frac{1}{1+x_{0}^2}|=|\frac{x_{0}^2-x^2}{(1+x^2)(1+x_{0}^2)}|=\frac{|x_{0}-x||x_{0}+x|}{(1+x^2)(1+x_{0}^2)}\le \delta|x+x_{0}|$

The last line uses the fact that $x^2, x_{0}^2\ge 0$.  How can I finish the proof?
 [/INST] Write 
$$|f(x)-f(y)|=\left|\frac 1{1+x^2}-\frac 1{1+y^2}\right|=\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|\leq |x-y|\frac{|x|+|y|}{(1+x^2)(1+y^2)}\\\
\leq \frac{|x-y|}2\frac{x^2+1+y^2+1}{(1+x^2)(1+y^2)}\leq \frac{|x-y|}2\left(\frac 1{1+y^2}+\frac 1{1+x^2}\right)\leq |x-y|$$
and now the $\delta$ you have to choose is clear. In fact, $f$ is lipschitz continuous.
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