Fitting an infinite collection of balls in an infinite dimensional unit ball

Given an infinite dimensional normed linear space, how would one show that it is possible to fit an infinite collection of non-overlapping balls of radius 1/4 in the unit ball?

I guess one can immediately reduce the problem to a normed linear space of countably infinite dimensions. The solution seems clear if the concept of orthogonality exists, but not every normed linear space has an inner product so it's not possible to apply something like Gram-Schmidt to produce an orthogonal basis. Is there any way around this, or there another approach that can be used?

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Here is a standard lemma you could use. If $M$ is a closed proper subspace of a normed linear space $E$, then for all $\epsilon\gt0$ there is an $x\in E$ of norm 1 whose distance to $M$ is greater than $1-\epsilon$. (E.g., here's a proof in Lemma 3-6.10 of Tsoy-Wo Ma's Classical analysis on normed spaces.)
Here's how you could use it. Let $B$ denote the unit ball of an infinite dimensional normed space $X$. Let $x_1\in X$ have norm 1, and let $M_1$ be the span of $\{x_1\}$. By the lemma there is an $x_2\in X$ of norm 1 whose distance to $M_1$ is greater than $\frac{2}{3}$. Let $M_2$ be the span of $\{x_1,x_2\}$, and let $x_3\in X$ have norm 1 and distance greater than $\frac{2}{3}$ to $M_2$. Repeat countably infinitely many times to obtain a sequence $x_1,x_2,\ldots$ of elements of $X$ of norm 1 with pairwise distances greater than $\frac{2}{3}$. Note that the lemma will always apply, because each $M_k=\operatorname{span}\{x_1,x_2,\ldots,x_k\}$ is finite dimensional, hence closed and proper. Then the balls of radius $\frac{1}{4}$ centered at the points $\frac{3}{4}x_1,\frac{3}{4}x_2,\ldots$ are disjoint and contained in $B$.
This works for any radius $r$ less than $\frac{1}{3}$, by changing $\frac{2}{3}$ to $\frac{2r}{1-r}$ and $\frac{3}{4}$ to $1-r$. – Jonas Meyer Sep 16 '10 at 23:17
In a Hilbert space, you could take $r=\sqrt2-1$ using an orthogonal sequence of vectors of length $2-\sqrt2$ for the centers. – Jonas Meyer Sep 16 '10 at 23:39