How to define Continuous $f: B\rightarrow B$ without fixed points? Let $X$ be a infinite dimensional Banach space and $B=\{x\in X:\ \|x\|\leq 1\}$.
How to define a continuous $f:B\rightarrow B$ without fixed points?
Edit 1: I have changed the question a litlle, because if $f$ is linear then $f(0)=0$.
 A: This paper shows that in an  arbitrary normed, infinite dimensional space, there is a continuous map on its unit ball (into its unit ball) that is fixed-point free with even stronger properties. 
According to the authors, the fact that every infinite dimensional normed space admits a fixed-point free, continuous map on its unit ball was first proved, using the Axiom of Choice, in Theorem 6.3 of [1], below, and was proven later  using "a more constructive proof", in [2], below. 
Of course, for specific spaces, it is not too hard to find fixed-free maps. For example, if $X=c_0$, then the map from $B_X$ to $B_X$ that sends the element $x=(x_1, x_2,\ldots)$ to $(1-\Vert x\Vert, x_1,x_2,\ldots)$ is continuous (in fact, non-expansive) and fixed-point free.  An example on the space $\ell_2$ is given in Patrick Da Silva's answer; and this can be generalized to examples for $\ell_p$ spaces as suggested by Tomás in the comments to your post.

[1] J. Dugundji, An extension of Tietze's theorem, Pacific Journal of math., I (1951), 353-367 
[2]  V. L. Klee, Jr., Some topological properties of convex sets, Trans. Amer. Math. Soc., 78, (1955), 30-45.
A: I think a good place to start would be if $X$ is an Hilbert space with a countable basis (yeah I know... lots of extras). This is because these spaces have a basis, hence we can write
$$
\mathcal E = \{ e_1, e_2, \dots, e_n, \dots \}
$$
as a Hilbert basis of the space. For $x \in B$, we have
$$
x = \sum_{n=1}^{\infty} (x,e_n) e_n = \sum_{n=1}^{\infty} \underset{(x,e_n)}{\underbrace{x_n}} e_n.
$$
Define
$$
f(x) \overset{def}{=} \frac 12 e_1 + \sum_{n=1}^{\infty} \frac{x_n}2 e_{n+1} = \frac 12 e_1 + \sum_{n=2}^{\infty} \frac{ x_{n-1} }2 e_n 
$$
It is continuous as the series part of the function is a linear bounded function and we simply translated it. (The coefficients $\frac 12$ are there to preserve $\| x \| \le 1$. I just realized I wrote all this answer assuming that this would be sufficient for $f(x) \in B$, but I now see that it is not. I guess this answer restricts to the case where $f(x)$, as defined above, is in $B$.) It has no fixed point, for if $f(x) = x$, we must have $x_{n-1} = x_n = 1/2$ by induction, which means $x \notin H$ because the norm of this candidate fixed point $x$ is not finite.
Now that the idea is there, we can go to Hilbert spaces with arbitrary (infinite) basis : choose a non-surjective injection
$$
\omega : \mathcal E \to \omega(\mathcal E)
$$
which has no finite cycles (and in particular, no fixed points). I believe this requires the axiom of choice at some point. Afterwards, letting
$$
x = \sum_{e \in \mathcal E} (x,e) e = \sum_{e \in \mathcal E} x_e e,
$$
let $e_0 \in \mathcal E \backslash \omega(\mathcal E)$ and define
$$
f(x) = \frac 12 e_0 + \sum_{e \in \mathcal E} x_e \omega(e).
$$
If $f(x) = x$, we have $x_e = x_{\omega(e)}$. If there exists an $e$ with $x_e \neq 0$, by assumption that $\omega$ has no finite cycles, this means
$$
x_e = x_{\omega(e)} = x_{\omega(\omega(e))} = \dots = x_{\omega^n(e)} 
$$
for every integer $n$, which leads to a contradiction because $x$ must have "bounded norm". Therefore the only possible fixed point left would be $x = 0$, but it is not fixed. Note that the same comment about $f(x) \in B$ applies here...
I believe one way to solve the problem of $f(x) \in B$ would be to replace my series on the right of my definition of $f(x)$ by a linear functional with no fixed point other than $0$ and with operator norm smaller than $\frac 12$ which is not surjective (so that you can find $e_0$). As for the general $X$ Banach space, I have no idea, but this is as close as I could get. My answer was meant to be a big comment.
Hope that helps,
