The result is a little more general. The postulated multiplication of sequences made of entries $\pm 1$ endows their $\{-1, 1\}^n$ set (having $N = 2^n$ cardinality) with the direct product group structure given by $(\{-1, 1\}, \cdot)$ (isomorphic with $\mathbb{Z}_2$). We will more generally prove Theorem. Given a finite abelian group $G$ of order $N$, and a subset $\emptyset \neq S \subseteq G$ of cardinality $k$, there exists an element $x \in G$ such that the intersection of $S$ with the set $xS = \{xs \mid s \in S\}$ has cardinality at most $\dfrac {k^2} {N}$. Proof. Consider the family $\mathcal{F}$ of the $N$ subsets $gS$ for all $g \in G$. It will be enough to prove there exist $g,h \in G$, $g\neq h$, such that $|gS \cap hS| \leq \dfrac {k^2} {N}$, since $|gS \cap hS| = |S \cap xS|$ for $x = g^{-1}h$. We have $|gS| = k$ for all $g \in G$. Moreover, any element $x \in G$ belongs to exactly $k$ of the sets of $\mathcal{F}$, since $x \in gS$ if and only if $g^{-1} \in x^{-1}S$. Therefore $\binom {k} {2} N = \sum_{g\neq h} |gS \cap hS|$, by double counting the elements of the pairwise intersections of the elements of $\mathcal{F}$. But $\binom {k} {2} N = \dfrac {k(k-1)} {2} N = \dfrac {N(N-1)} {2} \cdot \dfrac {k(k-1)} {N-1} \leq \dfrac {N(N-1)} {2} \cdot \dfrac {k^2} {N}$, since $k\leq N$. Since the sum $\sum_{g\neq h} |gS \cap hS|$ contains $\binom {N} {2} = \dfrac {N(N-1)} {2}$ terms, by an averaging argument, at least one of these terms has a value of at most $\dfrac {k^2} {N}$.

Perhaps an easier way is to use the probabilistic method. If we choose the sequence $C$ randomly as a vertex of the $n$-hypercube, it is clear that $\mathbb{E}[B\cap BC]=\frac{k^2}{2^n}$. This follows from expanding $B\cap BC$ as the sum of indicator variables and applying linearity of expectation. We’re then clearly done by the principles of the probabilistic method. Remark: This proves the general abelian group case as well!