$f(x)$ is uniformly continuous when $\lvert x_0 - x_1 \rvert \lt \delta \rightarrow \lvert f(x_0) - f(x_1) \rvert \lt \epsilon$ and $\delta$ itself depends upon $\epsilon$ only.
Lipschitz inequality defines a relationship between $\lvert x_0 - x_1 \rvert$ and $\lvert f(x_0) - f(x_1) \rvert$ such that there exists some constant $K$ and $\lvert f(x_0) - f(x_1) \rvert \lt K \times \lvert x_0 - x_1 \rvert = \epsilon$.
But there are relationships different from linear: for example, exponential. Thus, as far as I guess, $f(x)$ could be uniformly continuous indeed, however it is impossible to find corresponding constant $K$ valid for each and every possible $x_0$?
In other words, could some $f(x)$ be u.c., but dissatisfy given inequality at the same time?
You seem to be asking if being uniformly continuous implies Lipschitz continuity. This is false, as $$ f(x)=\sqrt{x} $$ is uniformly continuous on $[0,1]$ but not Lipschitz.
Suppose it were. Then we would have the following inequality for any $0\leq x<y\leq 1$ for some $K\in \mathbb{R}$. $$ |\sqrt{x}-\sqrt{y}|\leq K|x-y| $$ but then by the mean value theorem, this would mean that for some $x<c<y$, $$ \left|\frac{1}{2\sqrt{c}}\right|\leq K|x-y| $$ Now send $y\to 0$ and see that this would imply that $$ +\infty\leq 0 $$