I tried to do it like this:
Let for any map $g : \Delta \to R$ we have one element $b \in R$ that satisfies following: for every $\varepsilon \in \Delta$, the equality $g(\varepsilon)=g(0)+b\varepsilon$ holds.
Let's define map $d: \Delta \to R$. Let's suppose that $d$ is defined for every $\varepsilon \in \Delta$ but $d$ is discontinuous. We get that equality $d(\varepsilon)=d(0)+b\varepsilon$ doesn't hold because there is no such element $b$ that satisfies this equality.
Now, I don't know how to conclude from this statement that $d$ isn't defined for all $\varepsilon \in \Delta$.
How to prove that we can't define discontinuous function on $R$?
Thanks.
If $a$ and $b$ are in $R$ such that $a-b=\epsilon$ is in $\Delta$, then we have $f(b)=f(a) + k \epsilon$ for a suitable slope $k$, as you mentioned. But $k\epsilon$ is necessarily in $\Delta$, as well. Therefore the function is continuous. One way of summarizing this is that $\Delta$ is an ideal in $R$ but this claim should be checked (there might be some subtleties because the background logic is intuitionstic; I haven't thought this through fully).