Remind that $n$-dimensional binary cube (boolean) is a graph with vertices labeled by binary sequences of length $n$, and there is an edge between any two vertices if and only if their sequences differ only in one corresponding coordinate (digit).

Then our problem is equivalent to the following: prove that if from a $10$-dimensional binary cube we remove $8$ edges, then we always obtain a hamiltonian graph (i.e., a graph having a hamiltonian cycle – a closed path passing through all vertices of the graph exactly once).

By induction on $n$ let's prove a more general statement:
If from an $n$-dimensional binary cube, $n > 1$, we remove arbitrary $(n-2)$ edges, then the obtained graph is hamiltonian.

Proof: For $n=2$ the statement is obvious, there is nothing to remove.

Assume that for $n = k > 1$ the statement is true, and prove it for $n = k + 1$.

Let $V = \{(a_0, a_1, \dots, a_k) : a_i \in \{0, 1\}, 0 \le i \le k\}$ be the set of vertices of the $(k+1)$-dimensional cube $C = \langle V, E \rangle$, where $E$ is a set of edges. Let's take an arbitrary removed edge $e \in E$. W.l.o.g. we may assume that it connects vertices $u = (0, b_1, \dots, b_k)$ and $v = (1, b_1, \dots, b_k)$.

Then consider the sets of vertices
$$V^0 = \{(0, a_1, \dots, a_k) : a_i \in \{0, 1\}, 1 \le i \le k\}$$
and
$$V^1 = \{(1, a_1, \dots, a_k) : a_i \in \{0, 1\}, 1 \le i \le k\},$$
which form a partition of $V$; and corresponding subgraphs $C^0 = \langle V^0, E \setminus V^0 \rangle$ and $C^1 = \langle V^1, E \setminus V^1 \rangle$.

Note that:
1) Subgraphs $C^0$ and $C^1$ are identical to $k$-dimensional binary cubes;
2) A mapping $(0, a_1, \dots, a_k) \to^f (1, a_1, \dots, a_k)$ is an isomorphism from $C^0$ to $C^1$.

Let $D \subset E$ be the set of all removed edges, $|D| = n - 2 = k - 1$. Then we have a partition $D = D_0 \cup D_1 \cup D_2$, where $D_i$ is the set of edges removed from subgraph $C^i$, $i = 0, 1$, and $D_2$ is the set of removed edges which "connect" $C^0$ and $C^1$. In particular, $e \in D_2$.

Since $|f(D_0) \cup D_1| \le k - 2$, then by the induction hypothesis in $C^1$ there is a hamiltonian cycle $v = v_1, v_2, \dots, v_{2^k}$, which does not use the edges from $f(D_0) \cup D_1$. Then $u = u_1, u_2, \dots, u_{2^k}$ is a hamiltonian cycle in $C^0$, where $f(u_i) = v_i$ for $1 \le i \le 2^k$, not passing through the edges of $D_0$.

$$u = u_1, u_2, \dots, u_i, v_i, v_{i-1}, \dots, v_1, v_{2^k}, v_{2^k-1}, \dots, v_{i+1}, u_{i+1}, \dots, u_{2^k}$$
is a hamiltonian cycle in $C$, not passing through the edges of $D$. The statement is proved. $\square$