Addition is continuous with regard to a topology on the space of continuous functions on an open interval This is question $1.23$ in Rudin's "Functional Analysis." Let $X$ be a vector space of all continuous real valued functions on $(0,1)$. For each $r>0$ and $f\in X$ set $$V(f,r):=\{g\in X:|f(x)-g(x)|<r~\text{for all}~x\in(0,1)\}.$$
Let $\tau$ be the topology generated by these sets. I.e $\tau$ is the collection of all unions of finite intersections of the $V$'s. I want to prove that addition is continuous with respect to $\tau$. Let $U$ be a neighbourhood of $f+g$ for $f,g\in X$. Let us assume for the moment that $U=V(h,r)$, as if we can prove what is needed for this case, the case where $U$ is a finite intersection of $V$'s should follow easily enough. 
We want to find neighbourhoods $V_f$ and $V_g$ of $f$ and $g$ respectively such that $V_f+V_g\subset U$. For any $\varepsilon>0$ we get that $V(f,\varepsilon)$ and $V(g,\varepsilon)$ are neighbourhoods of $f$ and $g$ respectively. We want to show that for any $\phi=\phi_f+\phi_g\in V_f+V_g$ that $|(\phi-h)(x)|<r$ for all $x\in (0,1)$. Now we can show for all $x\in (0,1)$ that
$$|(\phi-h)(x)|\leq |(\phi_f-f)(x)|+|(\phi_g-g)(x)|+|(f+g-h)(x)|<2\varepsilon+\sup_{x\in(0,1)}|(f+g-h)(x)|.$$
However, I do not believe this is enough, because $(0,1)$ is not compact so $f+g-h$ may not attain a maximum on this interval. This means that it is possible that $\sup_{x\in(0,1)}|(f+g-h)(x)|=r$, which means that we can't get the sharp bound we need.
Any comments on how to modify my proof to take this possibility into account would be greatly appreciated. I would not be opposed to completely different approaches either.
 A: As you found, it is in general impossible to find neighborhoods $V(f,r_1)$ and $V(g,r_2)$ such that their sum is contained in a given neighborhood $V(h,r)$ containing $f+g$. So, working with $V$-type neighborhoods is not convenient in this problem. Consider the general form of neighborhoods, 
$$
U = \bigcap_{k=1}^n V(f_k, r_k)
\tag1$$ 
Observe that $U$ consists of all continuous functions $f$ such that 
$$
\max_k(f_k-r_k) < f < \min_k( f_k + r_k)\quad \text{ on } (0,1)
$$
Such a  neighborhood can be described as $U(v,w): = \{f : v<f<w \}$ where $v,w$ are continuous, $v<w$ on $(0,1)$, and $\sup(w-v)<\infty$. Conversely, any such $U(v,w)$ can be written in the form (1);  indeed, 
$$U(v,w) = V(v+R, R)\cap V(w-R, R)$$
for any $R$ such that $R>\sup(w-v)$.
So, consider a neighborhood $U(v,w)$ containing $f+g$. We have 
$v<f+g<w$. It follows that 
$$\frac{v+f-g}{2}<f<\frac{w+f-g}{2}$$ 
$$\frac{v+g-f}{2} <g< \frac{w+g-f}{2}$$  For any $h_1\in U\left(\frac{v+f-g}{2}, \frac{w+f-g}{2}\right)$ and $h_2\in U\left(\frac{v+g-f}{2} , \frac{w+g-f}{2}\right)$ we have 
$$\frac{v+f-g}{2} + \frac{v+g-f}{2} < h_1+h_2 < \frac{w+f-g}{2} +\frac{w+g-f}{2}$$
hence $h_1+h_2\in U(v,w)$ as desired.
