I've done some exercises on continuity using the $\varepsilon$-$\delta$ definition. Now going through my script I saw that there is a stricter version of continuity called uniform continuity. After watching some videos I think I understand the difference and the importance of the quantifiers and their order. I've seen many questions here that ask about the intuition behind continuity, but I could not find the difference in proving it.
My question now is: Is the following statement true? Apologies for the informal way of writing this.
When working through an exercise and showing that a function is continuous, we take $|f(x)-f(x_{0})|<\varepsilon$ and then we rearrange things until we find $|x-x_{0}|$. From the definition we know that this is $<\delta$ and we start to "estimate" and calculate some things until we get something of the form: $\delta = \varepsilon\ldots$
In the proof, we then plug it in to "hopefully" get rid of everything and only $\varepsilon$ remains.
Is it true that if only $\varepsilon$ remains, the function is uniformly continuous since it does not depend on the $x_{0}$? And that if the $x_{0}$ were to remain, we have that it is continuous, but not uniformly?
If you are able to find a $\delta>0$ which depends on $\varepsilon$ but not on $x_0$ then, yes, you shall have proved that $f$ is uniformly continuous.
But what if every $\delta$ that you are able to find depends upon $\varepsilon$ and upon $x_0$? Then you shall have proved that $f$ is continuous, but don't jump to the conclusion that it is not uniformly continuous. Take, for instance$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\sqrt x.\end{array}$$If $x_0\in[0,1]$, then$$x-x_0=\left(\sqrt x-\sqrt{x_0}\right)\left(\sqrt x+\sqrt{x_0}\right)$$and therefore\begin{align}\left|f(x)-f(x_0)\right|&=\left|\sqrt x-\sqrt{x_0}\right|\\&=\frac{\left|x-x_0\right|}{\sqrt x+\sqrt{x_0}}\\&\leqslant\frac{\left|x-x_0\right|}{\sqrt{x_0}}\end{align}and therefore you can take $\delta=\sqrt{x_0}\,\varepsilon$. But this only works if $x_0\ne0$ and you shall have to find another $\delta$ for the case in which $x_0=0$ ($\delta=\varepsilon^2$ will do then). So, the $\delta$ does depend upon $\varepsilon$ and upon $x_0$. However, $f$ is uniformly continuous (as is every continuous function form an interval $[a,b]$ into $\Bbb R$). The fact that this approach leads to a $\delta$ which depends upon both $\varepsilon$ and $x_0$ doesn't prove that there is no $\delta$ which depends only upon $\varepsilon$.