No. I believe the first counterexample is from:
Kinoshita, S. On Some Contractible Continua without Fixed Point Property. Fund. Math. 40 (1953), 96-98
which I unfortunately can't find online. Kinoshita's example is described on page 127 in this excellent article by Bing, however.
EDIT: The question has been revised to add the local connectivity condition; as stated, I think the question is open. If "contractible" is replaced with "acyclic" there are counterexamples dating back to Borsuk, referenced e.g. here; Borsuk's paper, which I can't find online, is:
K. Borsuk, Sur un continua acyclique qui se laisse transformer topologiquement en lui-meme sans points invariants, Fund. Math. 24 (1935), pp. 51-58.
This suggests that if the OP's conjecture is true, it is unlikely to be open to homological attacks.
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.
Best Answer
The theorem is equivalent to the determinacy of the Hex game. That's a very famous `application'.
The details can be found in [David Gale (1979). "The Game of Hex and Brouwer Fixed-Point Theorem". The American Mathematical Monthly 86: 818–827], a beautiful paper, which JSTOR serves at http://www.jstor.org/stable/2320146.