Let $X$ be a Tychonoff topological space and let $x\in X$. Let $B\subset C(X)$ be convex and compact in the topology of pointwise convergence, and such that $f(x)=1$, for every $f\in B$.
Is there an open neighborhood $U$ of $x$ such that $f(y)\ne 0$, for every $f\in B$ and $y\in U$?
Since $C(X)$ is not complete one cannot take the closed convex hull of the example in the comment. But what about this:
Let $g_n=1-f_n$ with $f_n$ as in my comment. Since the $g_n$ are bounded by one, the linear map $T:\ell^1\to C(X)$, $\lambda\mapsto \sum\limits_{n=1}^\infty \lambda_ng_n$ is well defined. We would like to have $T$ continuous as a map $(\ell^1,\sigma(\ell^1,c_0)) \to (C(X),pw)$ (where $c_0$ is the space of all null sequences so that $\ell^1$ is its dual). This follows from the pointwise convergence to $0$ of all sequences $(g_n(x))_{n\in\mathbb N}$. The unit ball $K$ of $\ell^1$ is weak$^*$ compact à la Alaoglu and hence $T(K)$ is compact and convex in $(C(X),pw)$. Then $B=\{1-g: g\in T(K)\}$ is convex and compact in the pointwise topology, it satisfies $f(0)=1$ for all $f\in B$ (because $g_n(0)=0$ for all $n$), and it contains all $f_n=1-g_n=1-T(e_n)$ with the standard unit vectors $e_n$ of $\ell^1$.
