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.
Usually it uses the fact:
If $a < s < t < u < b$ then $$\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.$$
I wonder whether any other version of this proof exists or not?