Let $S$ be an uncountable set and consider the space $\{0,1\}^S$ under the product topology where $\{0,1\}$ is discrete. Let $f:\{0,1\}^S\to[0,1]$ be a continuous function and suppose $f(\sigma)=1$ for some $\sigma\in\{0,1\}^S$. I need to show that $f(\tau)=1$ for some $\tau\in\{0,1\}^S\setminus\{\sigma\}$.
My first instinct was to take note of the cardinalities. Since $S$ is uncountable, $\{0,1\}^S$ has a cardinality of at least $2^{\aleph_1}$ whereas $[0,1]$ has a cardinality of only $2^{\aleph_0}$, so $f$ cannot be injective. Furthermore, every nonempty basis element in the product topology is a product of subsets of $\{0,1\}$ uncountably many of which must be $\{0,1\}$. So, every nonempty basis element, and therefore every nonempty open set in $\{0,1\}^S$, must have a cardinality of at least $2^{\aleph_1}$.
Since $f$ is continuous, $(a,1]$ is open in $[0,1]$, and $f(\sigma)=1$, $\ f^{-1}(a,1]$ must have a cardinality of at least $2^{\aleph_1}$, so $f$ cannot be one-to-one on any interval containing $1$. In fact, for any interval $(a,1]$, there must exist $b\in(a,1]$ such that $f^{-1}(\{b\})$ is uncountable. My problem is that I have no idea how to use this to conclude that $f^{-1}(\{1\})$ contains more than just $\sigma$.
Another potentially useful thing is that $\{0,1\}^S$ is compact since it is a product of compact sets which means the Extreme Value Theorem applies here.
Note that $f^{-1}(\{1\}) = \cap_{n>0}\, f^{-1}\left((1-n^{-1}, 1]\right)$. In particular $\sigma$ is contained in a countable intersection of basis open sets. Conclude that there is an enumerable set $T\subset S$ such that $f(\tau) = f(\sigma)$ if $\tau$ and $\sigma$ coincide on $T$.