The Lawvere fixed point theorem asserts that if $X, Y$ are objects in a category with finite products such that the exponential $Y^X$ exists, and if $f : X \to Y^X$ is a morphism which is surjective on points in the sense that the induced map $\text{Hom}(1, X) \to \text{Hom}(1, Y^X)$ is surjective, then $Y$ has the fixed point property: for every morphism $g : Y \to Y$ there exists a point $y : 1 \to Y$ such that $g \circ y = y$.
The Brouwer fixed point theorem asserts that the closed $n$-disks, all of which I will denote by $D$ for ease of notation, have the fixed point property as objects of $\text{Top}$.
Seeing these two theorems together, it is tempting to try to prove the latter from the former by finding a topological space $X$ such that the exponential $D^X$ exists, together with a surjective continuous map $X \to D^X$. Does there in fact exist such an $X$?
Edit, 4/13/17: I'm still interested in this question, and so are some people associated with MIRI (at least when $n = 1$); for some details about why see here.
Best Answer
In my experience it is worth considering variants of Lawvere fixed point theorem. In the present case, I would split things up as follows, in order to circumvent the non-constructive nature of Brouwer's fixed point theorem.
Also, let me point out that we need not worry about exponentials too much, even though they do not exist in the category of topological spaces, unless the exponent is nice enough. We can move over to a cartesian-closed subcategory, such as teh compactly generated spaces, or to a cartesian closed supcategory, such as equilogical spaces.
Theorem: [Approximate Lawvere] Suppose $(B, d)$ is a metric space and $e : A \to B^A$ is a continuous map, such that for every continuous map $g : A \to B$ and $\epsilon > 0$ there is $a \in A$ such that $d(e(a)(a), g(a)) < \epsilon$. Then every continuous map $f : B \to B$ has approximate fixed points: for every $\epsilon > 0$ there is $b \in B$ such that $d(b, f(b)) < \epsilon$.
Proof. Given any $f$ and $\epsilon$ consider the map $g(a) = f(e(a)(a))$. there is $a \in A$ such that $d(e(a)(a), g(a)) < \epsilon$ and then $b = e(a)(a)$is an $\epsilon$-approximate fixed point of $f$. QED.
One way to use the theorem is via the sup metric (allowing infinite distance):
Corollary: If $e : A \to B^A$ has a dense image in the sup metric on $B^A$ then every endomap on $B$ has approximate fixed points.
Suppose we could apply the previous theorem to the closed ball $D^n$. Then we would know (constructively!) that every endomap on $D^n$ has approximate fixed points. then we just have another easy step, which contains all the classical reasoning needed:
Theorem: Suppose $X$ is compact and $f : X \to X$ has an $\epsilon$-approximate fixed point for every $\epsilon > 0$. Then $f$ has a fixed point.
Proof. By countable choice, for every $n$ there is $x_n \in X$ such that $d(x_n, f(x_n)) < 1/n$. Because $X$ is compact, $x_n$ has a subsequence converging to some $y \in X$. It is now easy to see that $y$ is a fixed point of $f$. QED.
But I do not see how to apply the approximate Lawvere to the closed ball, if that is even possible.