A convex real function is continuous - can we generalize? There is a well-known theorem $\newcommand{\RR}{\mathbb{R}}$

Let $f: (a,b) \rightarrow \RR$ be a convex function. Then $f$ is continuous on $(a,b)$

The proof I know makes use of the fact that the real numbers form an ordered field and uses the differential quotient to show that $f$ is Lipschitz continuous. So the method cannot be used to prove the following fact (if it's true at all):

Let $A$ be a convex set. Let $f: A \rightarrow \RR$ be a convex function. Then $f$ is continuous on $\operatorname{Int} A$

Is my proposition true at all? If yes, how can we prove it? If no, please provide a counterexample.
 A: It's wrong in infinte dimensions, where we have even discontinuous linear functions $f \colon X \to \mathbf R$ on every infinite-dimensional normed space $X$. But one can say the following:

Proposition 1. Let $X$ a normed space, $B_X$ its open unit ball, $x \in X$, $r > 0$, $0 < s < r$. Let $f \colon x + rB_X \to \mathbf R$ be a convex function. Then
(a) If $f \le m$, then $\def\abs#1{\left|#1\right|}\abs f \le \abs m + 2\abs{f(x)}$.
(b) If $\abs f \le m$, then $f$ is $\frac{2m}s$-Lipschitz on $x + (r-s)B_X$.

Proof. Without loss of generality (just simplifying the notation), we assume that $x = 0$. 
(a) For $y \in rB_X$, we have by convexity 
$$ f(0) \le \frac 12 f(y) + \frac 12f(-y)$$
hence 
$$ f(y) \ge 2f(0) - f(-y) \ge 2f(0) - m $$
and this implies 
$$ \abs{f(y)} \le \max\{\abs m, 2\abs{f(0)} + \abs{m}\} \le \abs{m} + 2\abs{f(0)} $$
(b) Let $y,z \in (r-s)B_X$. Define 
$$ a := y + \frac{s}{\def\norm#1{\left\|#1\right\|}\norm{y-z}}(z-y) \in rB_X $$
Then $y$ is a convex combination of $a$ and $z$, namely
$$ y = \frac{s}{s + \norm{y-z}}z + \frac{\norm{y-z}}{s + \norm{y-z}}a $$
Convexity of $f$ implies 
$$ (s + \norm{y-z})f(y) \le sf(z) + \norm{y-z}f(a) $$
Thus 
$$ s\bigl(f(y) - f(z)\bigr) \le \bigl(f(a)-f(y)\bigr)\norm{y-z} \le 2m\norm{y-z} $$
By exchanging the roles of $y$ and $z$, we get along the same lines 
$$ s\bigl(f(z) - f(y)\bigr) \le \bigl(f(a)-f(y)\bigr)\norm{y-z} \le 2m\norm{y-z} $$
Together this implies 
$$ \abs{f(z) - f(y)} \le \frac{2M}s \norm{z-y} $$
and $f$ is Lipschitz on $(r-s)B_X$. $\square$

From this we get

Proposition 2. Let $X$ be a normed space, $C \subseteq X$ be open and convex, $f \colon C \to \mathbf R$ convex. Then the following are equivalent:
  (a) $f$ is locally Lipschitz.
  (b) $f$ is continuous.
  (c) $f$ is locally bounded.
  (d) $f$ is upper bounded on a non-empty open subset of $C$.

Proof. Obviously, $(a) \def\To{\Rightarrow}\To (b) \To (c) \To (d)$, we it remains to prove that $(d) \To (a)$. There is an open ball, $x + rB_X \subseteq C$ on which $f \le m$ for some $m \in \mathbf R$. Then by the above, $\abs f$ is bounded by $\abs m + 2\abs{f(x)}$, and hence is Lipschitz on $x + \frac r2B_X$. That is $f$ is locally Lipschitz. $\square$
In infinite dimensions, that is all what can be said. In finite dimensions, we have

Corollary. Let $A \subseteq \mathbf R^n$ be convex, $f \colon A \to \mathbf R$ a convex function. Then $f$ is continuous on $\def\i{\operatorname{int}}\i A$. 

Proof. Let $C := \i A$. Then $C$ is convex and $f\colon C \to \mathbf R$ 
is convex. Let $x \in C$, as $C$ is open, there are $x_1, \ldots, x_k$ with
$x \in \i \mathrm{conv}\{x_1, \ldots, x_k\}=: U$ as there is same $n$-cube around $x$ with $x$ as a center. As $f$ is convex, $f \le \max\{f(x_1), \ldots, f(x_k)\}$ on $U$. Therefore, by Proposition 2, $f$ is continuous on $C$. $\square$
