I have following proof to this theorem: (Source)
A real-valued function $f$ defined in $(a,b)$ is said to be convex if $$f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(y)$$ whenever $a<x<b$, $a<y<b$, $0<\lambda<1$. Prove that every convex function is continuous.
Proof: Fix points $c,d$ with $a<c<d<b$, let $\eta>0$ be any fixed positive number with $\eta<\dfrac{d-c}2$ and consider any two points satisfying $c+\eta\le x<y\le d-\eta$. Then it is showed that there there exists $M$ with $|f(t)|<M$ for all $t\in [c,d]$.
Now, we can write $$x=(1-\lambda)c+\lambda y,$$ where $\lambda=\dfrac{x-y}{y-c}\in(0,1)$. Accordingly we have $$f(x)-f(y)\le(1-\lambda)(f(c)-f(y))=\dfrac{y-x}{y-c}(f(c)-f(y))\le\dfrac{y-x}{\eta}|f(c)-f(y)|.$$ Thus, $$f(x)-f(y)\le\dfrac{2M}{\eta}(y-x).$$ Similarly, $$f(y)-f(x)\le\dfrac{2M}{\eta}(y-x).$$ Therefore $$|f(y)-f(x)|\le\dfrac{2M}{\eta}|y-x|$$ for all $x,y\in [c+\eta,d-\eta]$. Since $c,d$ and $\eta$ are arbitrary, it follows that $f$ is continuous on $(a,b)$.
1. I can't follow this last line: I think that bound $M$ depends on choice of $c,d$, so how can we be sure that there exists $M_0$ such that $$|f(y)-f(x)|\le\dfrac{2M_0}{\eta}|y-x|$$ for all $x,y\in (a,b)$?
2. What goes wrong if domain is not open interval?
1. I agree with the comment above. For any $ x $ and $ y $ you want to analyze, we can always find such $c$ and $d$ and fix them and find such $M$ as shown above.
2. If the domain is not open, e.g. $[a,b]$, we can choose $f(a)$ arbitrarily as long as $f(a)>\lim_{x\to a^+}$, so $f$ is not continuous at $a$.