Prove increasing convex function has increasing differences

Since $x_1 \le x_1+t,x_2 \le x_2+t$, from the definition of convexity we have $$ v(x_1+t) \le \left(\frac{x_2-x_1}{x_2-x_1+t}\right)v(x_1) + \left(\frac{t}{x_2-x_1+t}\right)v(x_2+t) $$ and $$ v(x_2) \le \left(\frac{t}{x_2-x_1+t}\right)v(x_1) + \left(\frac{x_2-x_1}{x_2-x_1+t}\right)v(x_2+t).$$ Adding up the inequalities gives the desired inequality.

[This inequality is true for any convex function, increasing or not.]

I'm going to assume that $x_2\le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1\le x_2\le x_1+t$. We have $$ x_2={x_1+t-x_2\over t}x_1+{x_2-x_1\over t}(x_1+t), $$ so by the convexity inequality for $\nu$, $$ \nu(x_2)\le {x_1+t-x_2\over t}\nu(x_1)+{x_2-x_1\over t}\nu(x_1+t). $$ Similarly, $$ \nu(x_1+t)\le {x_2-x_1\over t}\nu(x_2)+{x_1-x_2+t\over t}\nu(x_2+t). $$ Now add these two inequalities, clear $t$ from the denominator, and simplify.