I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$
Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.
Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.
Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.