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 $B$.