Let $f(x)=\frac{x^2}{x^2+1}$ where $f:\mathbb{R} \to \mathbb{R}$ examine uniform continuity and lipschitz condition.
I'm not sure if my solutuion is OK. $\displaystyle \ \ |f(x)-f(y)|= |\frac{(x-y)(x+y)}{(x^2+1)(y^2+1)}|< |x-y|$ since $\frac{x+y}{(x^2+1)(y^2+1)}<1$ so taking $\delta=\epsilon$ we will have $|f(x)-f(y)|<|x-y|<\epsilon$ so $f$ is uniform continuous.
To lipschitz condition we have $\displaystyle L\ge\frac{|f(x)-f(y)|}{|x-y|}$ taking $y=0, x=n$ we have $\displaystyle L \ge \frac{n^3}{n^2+1} \to \infty$ which is contradiction with $L< \infty $
It is the Lipschitz argument that is wrong: $\left | \frac{f(0)-f(n)}{n} \right | = \left | \frac{\frac{n^2}{n^2+1}}{n} \right | \to 0$ as $n \to \infty$. Here $f$ is indeed Lipschitz. An easier way to prove this is to check that the derivative is bounded, then apply the mean value theorem.