An approximate infinite-dimensional fixed point theorem Given $\epsilon > 0$ and $f : [0, 1]^{\omega} \rightarrow [0, 1]^{\omega}$, can we find $x$ such that $x \in \textrm{Conv}\left( \left\{f(y) : ||y - x||_{\infty} < \epsilon\right\}\right)$? 
In finite dimensions this is straightforwardly equivalent to Brouwer's fixed point theorem. In the infinite dimensional case they are not equivalent, and I don't even know whether to expect the claim to be true or false (I was expecting it to be much more straightforward to settle). 
I would actually be happy to solve the problem for the simpler case of maps $f : [0, 1]^{\omega} \rightarrow \left\{0, 1\right\}^{\omega}$, since I'm imagining each coordinate of $f$ as returning the truth of a certain predicate applied to $x$. And we could just as well work with the discrete setting, of $f : \left\{0, 1, \ldots, m\right\}^{\omega} \rightarrow \left\{0, m\right\}^{\omega}$. 
I'm aware of this result, but it doesn't seem to help very much here.
Edit: I would actually be happy to find an $x$ which isn't separated from $\left\{ f(y) : ||y - x||_{\infty} < \epsilon\right\}$ by any hyperplane parallel to the coordinate axes. This seems like it should be much easier.
 A: In fact there is a Lipschitz map which has no approximate fixed point in this sense.
Take some Lipschitz retraction $f$ from the interior of the cube into its exterior, and take $\epsilon$ much smaller than $1 /$ the Lipschitz constant of $f$. Consider the map $g$ which sends $x$ to its reflection about the point $(1/2, 1/2, \ldots)$.
Then $g$ has no fixed point in my sense. Any point close to a boundary can't be a fixed point because $f$ fixes the boundaries (so $g$ moves them far). So if $x$ is a fixed point, it must be strictly in the interior (it has to be pretty far from the exterior because $f$ is Lipschitz). Since $f$'s image is the exterior, there is a coordinate $i$ with $f(x)_i \in [-1, -1 + \epsilon] \cup [1 - \epsilon, 1]$. By the Lipschitz property $f(y)_i$ must be in one of these two intervals for all $y$ in the $\epsilon$-neighborhood of $x$. But $x_i$ is in $(-1 + \epsilon, 1 + \epsilon)$, since $x$ was far in the interior of the cube. Contradiction.
A: In fact the property fails on any non-compact closed convex set $K$ of a Banach space : there exists a Lipschitz map $f:K\to K$ with no approximate fixed point, that is $\inf_{x\in K}\|f(x)-x\|=\delta > 0$ (See  Geometric Nonlinear Functional Analysis- Part 1 by Benyamini and Lindenstrauss, Thm 3.4 ). If we take $\epsilon:=\delta/\mathrm{lip}(f)$ there holds $\operatorname{co}f\big(B(x,\epsilon)\big)\subset  B\big(f(x), \delta\big)\not\ni x ,$ for any $x\in K$.
