real analysis - Showing that a function is uniformly continuous but not Lipschitz

If $g(x):= \sqrt x$ for $x \in [0,1]$, show that there does not exist a constant $K$ such that $|g(x)| \leq K|x|$ $\forall x \in [0,1]$

Conclude that the uniformly continuous function $g$ is not a Lipschitz function on interval $[0,1]$.

Necessary definitions:

Let $A \subseteq \Bbb R$. A function $f: A \to \Bbb R$ is uniformly continuous when:
Given $\epsilon > 0$ and $u \in A$ there is a $\delta(\epsilon, u) > 0$ such that $\forall x \in A$ and $|x - u| < \delta(\epsilon,u)$ $\implies$ $|f(x) - f(u)| < \epsilon$

A function $f$ is considered Lipschitz if $\exists$ a constant $K > 0$ such that $\forall x,u \in A$ $|f(x) - f(u)| \leq K|x-u|$.

Here is the beginning of my proof, I am having some difficulty showing that such a constant does not exist. Intuitively it makes sense however showing this geometrically evades me.

Proof (attempt):

Suppose $g(x): = \sqrt x$ for $x \ in [0,1]$
Assume $g(x)$ is Lipschitz. $g(x)$ Lipschitz $\implies$ $\exists$ constant $K > 0$ such that $|f(x) - f(u)| \leq K|x-u|$ $\forall x,u \in [0,1]$.

Evaluating geometrically:

$\frac{|f(x) - f(u)|}{ |x-u|}$ = $\frac{ \sqrt x - 1}{|x-u|}$ $\leq K$

I was hoping to assume the function is Lipschitz and encounter a contradiction however this is where I'm stuck.

Can anyone nudge me in the right direction?