Prove that a convex curve C has tangent lines everywhere except at countably many points. Today, I was faced with the following question:

Suppose E was a convex region in the plane bounded by a curve C. Show that C has a tangent line except at a countable number of points.

Since the next question asks me to show that every convex function $f:I\subset \mathbb{R}\rightarrow \mathbb{R}$ has a derivative everywhere except at a countable number of points, I am trying to avoid proving the first statement using derivatives. However, I have no idea even how to attempt this question because I don't know how I can characterize a curve that bounds a convex region. I have done some reading and it led me to a field I know nothing about so I am pretty much stuck here. The question gives the hint of considering circles, triangles and so on but I don't know where that is supposed to lead me because I can easily think of convex regions which are not polygons (or convex regions with 2 points without a tangent line, for example). Even when I tried to use derivatives to solve the problem I had no idea how to proceed, so I would greatly appreciate some help.
EDIT: After noticing that there's probably no way around using derivatives, I would like to point out that answers using derivatives are welcome too.
 A: The key concept in answering your question is the subdifferential of the function. Whilst a derivative of a convex function does not need to be defined at every point, its subdifferential is defined at every point. You can think of subdifferential $\partial f(x)$ as the closed interval between the left derivative $f'_-(x)$ of the function and the right derivative $f'_+(x)$ of the function $f(x)$. This interval is degenerate, i.e. consisting of a single point, when $f'_-(x)=f'_+(x)$, in that case $f$ is differentiable at $x$ (in other terms, it has a tangent).
A convex function has nonempty subdifferential at every interior point $x$ of $I$, so both left and right derivatives of $f$ are defined and $f'_-(x)\leq f'_+(x)$. What is more, the functions $f'_-(x)$ and $f'_+(x)$ are increasing thanks to the convexity of $f$. Consider the set $D$ of points $x$ at which $f'_-(x)\neq f'_+(x)$. Then the open intervals $(f'_-(x),f'_+(x)), x \in D$ are disjoint. Since each open interval contains a rational number and there are only countable many rational numbers, the set $D$ can be at most countable.
Let me clarify the last argument: Every $x\in D$ can be associated to an open interval $I(x)$ in a way that the intervals $I(x), x\in D$ are disjoint. Each of these intervals contains a rational number (in fact infinitely many of then, but that does not matter in this proof), pick any of them and denote it $r(x)$. So we have that $r(x)\in I(x)$ and $r(x)$ are different for each $x\in D$ as the intervals $I(x)$ are disjoint. Consider the set $R = \{r(x): x\in D\}$. Since it is a subset of $\mathbb Q$, the set $R$ is at most countable. Since $r:D\to R$ is a bijection between sets $D$ and $R$, $R$ is at most countable.
A good reference book would be Convex Analysis by R. Tyrrell Rockafellar for example.
