convex function in open interval is continuous

In any closed interval subset of the open interval, the function satisfies $$\displaystyle \frac{|f(x) - f(y)|}{|x-y|} \leq C$$

($\displaystyle C$ depending on the interval).

Intuitively, the slope of the line joining $(x, f(x))$ to $(y,f(y))$ increases if you move $x$ or $y$ to the right.

For each $x_0 \in C$, $f$ is both left- and right-differentiable at $x_0$, that is, $$ \mathop {\lim }\limits_{x \uparrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R} \;\; {\rm and} \;\; \mathop {\lim }\limits_{x \downarrow x_0 } \frac{{f(x) - f(x_0 )}}{{x - x_0 }} \in \mathbb{R}, $$ respectively. For a proof, see Theorem 14.5(a) on p. 248 of this book or Theorem 1(2) here. It follows that $f$ is both left- and right-continuous at $x_0$, hence continuous there.

Remark: A convex function on a closed interval need not be continuous at the end points (for example, the function $f$ on $[0,1]$ defined by $f(x)=x^2$ for $|x|<1$, $f(x)=2$ for $x \in \lbrace -1,1 \rbrace)$.

elaborating on another answer: for $x<y<z$ we have $$ \frac{f(y)-f(x)}{y-x}\leq\frac{f(z)-f(x)}{z-x} $$ employing convex combination (i.e. $y=tx+(1-t)z, t\in[0,1]$) $$ y=\frac{z-y}{z-x}x+\frac{y-x}{z-x}z $$ and the definition of convexity $$ f(tx+(1-t)y)\leq tf(x)+(1-t)f(y). $$