Lipschitz at every point in $[a,b]$ but not Lipschitz on $[a,b]$? The question is: 

Find a function on $[a,b]$ which is Lipschitz at every point of $[a,b]$, but not Lipschitz on $[a,b]$.

I am having a hard time picturing this, and thus, having a hard time finding such a function... I have not learned a lot about this other than the definition (not even derivatives yet!), so if someone can help me picture this that'd be awesome! 
 A: I assume that by “Lipschitz at every point” you mean: 
For every $x \in [a,b]$ there is $\varepsilon_x \gt 0$ and a constant $C_x$ (that may both depend on $x$) such that for all $y \in [a,b]$ with $|y-x| \lt \varepsilon_x$ we have  $|f(y) - f(x)| \lt C_x |y-x|$.
(I don't know if this is a standard notion, it's a definition I made up myself on the fly, because that's the only interpretation of “Lipschitz at every point” I could think of that 1. deserves the name, and 2. doesn't imply Lipschitz on a compact interval.)

For inspiration, the following picture may help: it is the graph of $x\sin{\frac{1}{x}}$. However, showing rigorously that it is Lipschitz at every point is a bit cumbersome without derivatives:

The picture should indicate what to do: make the function “oscillate” wildly so that the constants $C_x$ can't be chosen to be bounded above when $x$ ranges over $[a,b]$. This would almost be achieved by the function $\sin{\frac1x}$, but that function fails to be continuous (and hence Lipschitz) at $0$. That's why I added a factor $x$ to get $x\sin{\frac1x}$ and and thus squeeze the graph of $\sin{\frac1x}$ into the region $\{|y|\leq |x|\}$, thus ensuring that the function is Lipschitz at $0$.

For an example based on this idea that can be tackled without taking derivatives, I suggest the following construction on $[0,1]$:
Put $f(0) = 0$.
Divide the interval $[0,1]$ into the intervals $I_n = \left[\frac{1}{n+1},\frac{1}{n}\right]$, $n=1,2,3,\ldots$. 
If $n$ is odd define $f_n$ on $I_n$ to be the straight line segment that connects $\left(\frac{1}{n+1},-\frac1{n+1}\right)$ to $\left(\frac1n,\frac1n\right)$ and if $n$ is even define $f_n$ to be the straight line segment that connects $\left(\frac{1}{n+1},\frac1{n+1}\right)$ to $\left(\frac1n,-\frac1n\right)$.
The idea is: The length of the interval $I_n$ is $\frac{1}{n} -  \frac{1}{n+1} = \frac{1}{n(n+1)}$, while $|f(\frac{1}{n}) - f(\frac{1}{n+1})| = \frac{1}{n}+\frac{1}{n+1} = \frac{2n+1}{n(n+1)}$, so the function's slope on $I_n$ is $2n+1$. 
This will give you a function of the desired type which looks something like this:

I'll leave it to you to figure out the detailed formulas and to prove that it is Lipschitz at every point, but not Lipschitz on the entire interval $[0,1]$.
As soon as you have that, simply take $g(x) = f\left(\frac{x-a}{b-a}\right)$ on $[a,b]$.
