Suppose $\phi$ is convex in $(a,b)$. Then the slope $[\phi(x+h) - \phi(x)], h>0$, decreases with $h$. Hence, the derivative on the right,
$$D^+\phi(x) = \lim_{h \to 0+} \frac{[\phi(x+h) - \phi(x)]}{h},$$
exists and is distinct from $+\infty$ in $(a,b)$. Similarly, the derivative on the left, $D^-\phi(x)$ exists and is distinct from $-\infty$ in $(a,b)$. Since $[\phi(x) - \phi(x-h)]/h \le [\phi(x+h) - \phi(x)]/h, h>0$, we obtain
$$-\infty < D^- \phi(x) \le D^+\phi(x) < +\infty.$$
This shows in particular that $\phi$ is continuous in $(a,b)$.
How does this argument show that $\phi$ is continuous in $(a,b)$?
Existence of $\lim_{h\to 0^+} \frac{\phi(x+h)-\phi(x)}{h}$ imlplies that $$\lim_{h\to 0^+} h\cdot \frac{\phi(x+h)-\phi(x)}{h} = 0 \implies \lim_{y\to x^+} \phi(y) = \phi(x). $$ Similarly we would should the equality for left limit. But then, since left and right limits exist and are equal, then the limit itself must exist and be equal to them. So $$ \lim_{y\to x} \phi(y) = \phi(x).$$ This is just from the existence of the limits themself.