If $m: \mathbb R \times \mathbb R \to \mathbb R$ is continuous, I am having trouble seeing how $m^{-1}((0,1))$ is open in $\mathbb R \times \mathbb R$?
I know that $\{(x,y) \mid 0<x,y<1/2\}\subset m^{-1}((0,1))$ and $\{(x,y) \mid 1/2<x<1, 0<y<1/2\}\subset m^{-1}((0,1))$.
In general, $m^{-1}((0,1))=\{(x,y)\mid 0<x+y<1\}=\{(x,y)\mid -y<x<1-y\}$.
But what is the whole set $m^{-1}((0,1))$?
If $m: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is addition, so $m(x,y) = x+y$, then let $O \subseteq \mathbb{R}$ be open, and consider $m^{-1}[O]$. To see it is open we show that $(p,q) \in m^{-1}[O]$ is an interior point of it, so let $(p,q) \in m^{-1}[O]$, which means that $m(p,q) = p+q \in O$. As $O$ is open, there is some $\varepsilon>0$ such that the open interval $(p+q-\varepsilon, p+q+\varepsilon) \subseteq O$, by the (or a) definition of the topology of $\mathbb{R}$. Now, $$(p,q) \in U:=(p-\frac{\varepsilon}{2}, p+\frac{\varepsilon}{2}) \times (q-\frac{\varepsilon}{2}, q+\frac{\varepsilon}{2})$$ and $U$ is an open set of $\mathbb{R} \times \mathbb{R}$ in the product topology. It's clear from the ordered field properties of $\mathbb{R}$ that $U \subseteq m^{-1}[O]$, showing $(p,q)$ to be an interior point of $O$:
$(x,y) \in U$ iff $p - \frac{\varepsilon}{2} < x < p+\frac{\varepsilon}{2}$ and $q-\frac{\varepsilon}{2} < y < q+\frac{\varepsilon}{2}$ so that (adding these inequalities), we have $p+q -\varepsilon< x+y < p+q+\varepsilon$, so $m(x,y) \in O$ as required.
As we already know we have a group, for proving the group operation continuous it suffices it to be continuous at $(0,0)$ and so we could have limited ourselves to checking sets $m^{-1}[(-r,r)]$ to be open for all $r>0$, as such sets form a local base at $0$. This is entirely similar: just note that $(-\frac{r}{2}, \frac{r}{2}) \times (-\frac{r}{2},\frac{r}{2})$ will do as a neighbourhood of $(0,0)$ that maps to $(-r,r)$ under $m$, but slightly easier maybe.