Show $f$ is continuous on $(a,b)$, if $\forall C$ : closed subinterval, $\forall x,y : \in C, \exists M(C) >0 \ \ s.t. |f(x) - f(y) | < M(C) |x-y|$ Suppose the following holds, for $f : (a,b) \to \mathbb{R}$, and $\forall C$ : closed sub-interval in $(a,b)$,

$\forall x,y : x \neq y \in C, \exists M(C) >0 \ \ s.t. |f(x) - f(y) | < M(C) |x-y|$

Is it true that $f$ is continuous on $(a,b)$?

My try
Fix $x_0 \in (a,b)$ and pick $\epsilon >0$. 
In controlling the range of $x$ into $|x - x_0 | < \delta$, I thought that $\delta = \frac{\epsilon}{M(C)}$ might be appropriate, but since $M$ also depends on $x$, I think it was the wrong approach.
Can this statement be verified, or denied?

Fix $x \in (a,b)$. Then, there is $\delta_1 > 0$ such that $(x-2\delta_1, x+2\delta_1) \subset (a,b)$. Then, $[x-\delta_1, x+\delta_1] =: C \subset (a,b)$, so there is $M(C) > 0$ such that $|f(y) - f(z)| < M(C) |z-y|$ for all $z \neq y \in C$. In particular, $|f(y) - f(x)| < M(C) |y-x|$ for all $C \ni y \neq x$ Note that, since $\delta_1$ depends only on $x$, $C$ (hence $M(C)$ as well) depends only on $x$. Let $\epsilon > 0$ be arbitrary and take $\delta = \min\{ \epsilon/M(C), \delta_1 \}$. Then $|y-x| < \delta \implies |f(y) - f(x)| < \epsilon$. Again $\delta$ depends only on $x$ and $\epsilon$ because $M(C)$ and $\delta_1$ depend only on $x$. So $f$ is continuous (recall that, in the definition of continuity [not uniform continuity], $\delta$ is allowed to depend both on $x$ and $\epsilon$).
