Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.

You are nearly finishing the proof.

$$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$

Take $\delta=\epsilon$.

According to the mean value theorem, $$ \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a| $$ where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$. But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$. In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value. But that is not hard. So you have $$ |f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$}, $$ (where $M$ is the absolute maximum of $|f'|$). So $f$ is Lipschitz-continuous and therefore uniformly continuous.

Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq \frac{|x|}{1+x^2} < 1$$ and $$\frac{|a|}{(1+x^2)(1+a^2)} \leq \frac{|a|}{1+a^2} < 1$$