Let $n\in\mathbb{N}$ and consider $x,y \in\{0,1\}^n$. The Hamming distance of $x,y$ is defined by $$d_H(x,y) = |\{i\in \{0,\ldots, n-1\}:x_i\neq y_i\}|.$$
For $n\geq 2$ let $H(n,2)$ be the graph given by the vertex set $\{0,1\}^n$ and the edge set $$\big\{\{x,y\}: x\neq y\in \{0,1\}^n \land d_H(x,y) \leq 2\big\}.$$
Question. If $G=(V,E)$ is a finite graph, is there $n\in\mathbb{N}$ such that $G$ is isomorphic to an induced subgraph of $H(n,2)$?
Not always. Consider a graph with 9 vertices $1,2,\dotsc,9$, in which $1,2$ are not joined, but both $1,2$ are joined with $3,\dotsc,9$.
Assume that it is isomorphic to an induced subgraph of $H(n,2)$, with $x_i\in \{0,1\}^n$ corresponding to $i=1,2,\dotsc,9$. Then $x_1$ and $x_2$ differ in at least 3 positions. We may suppose that $x_1=(a_1,\dotsc,a_n)$, $x_2=(b_1,\dotsc,b_n)$ with $a_i\ne b_i$ for $i=1,2,3$. Let $i=3,4,\dotsc,9$. Then for $x_i=(c_1,c_2,\dotsc,c_{n})$ we get either $c_i\ne a_i$ for at least two indices $i\in \{1,2,3\}$, or $c_i\ne b_i$ for at least two indices $i\in \{1,2,3\}$. In the first case the inequality $d_H(x_i,x_1)\leqslant 2$ yields that $c_i\ne a_i$ for exactly two indices $i\in \{1,2,3\}$, and $c_i=a_i$ for $i>3$. There exist exactly 3 such sequences. And 3 more sequences in the second case. But we should have 7 distinct sequences $x_i$'s, $i=3,4,\dotsc,9$. A contradiction.
graphviz knowledge, I added a picture. If it doesn't help, then please feel free to delete it.
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