Proof of "every convex function is continuous" 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 $a &lt; x &lt; b,\; a &lt; y &lt; b,\; 0&lt; \lambda &lt;1$.
  Prove that every convex function is continuous.  


Usually it uses the fact:
If $a &lt; s &lt; t &lt; u &lt; b$ 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?
The pictorial version.  (But it is the same as your inequality version, actually.)

Suppose you want to prove continuity at $a$.  Choose points $b,c$ on either side.  (This fails at an endpoint, in fact the result itself fails at an endpoint.)



By convexity, the $c$ point is above the $a,b$ line, as shown:



Again, the $b$ point is above the $a,c$ line, as shown:



The graph lies inside the red region, 



so obviously we have continuity at $a$.
The best alternative proof (in my humble opinion) is a function is convex if and only if its epigraph is a convex set.  If a function is NOT continuous then the epigraph can't be convex (obviously... draw a picture); but then by the above, the function can't be convex.  This proof used the contrapositive.