Any orthonormal set extends to an orthonormal basis, over any field of characteristic not $2$. This is a special case of Witt's theorem.

Any orthonormal set extends to an orthonormal basis, over any field of characteristic not $2$. This is a special case of Witt's theorem.

EDIT: In response to Vipul's comment: The proof of Witt's theorem is constructive, and leads to the following recursive algorithm for extending an orthonormal set $\lbrace v_1,\ldots,v_r \rbrace$ to an orthonormal basis.

Let $e_1,\ldots,e_n$ be the standard basis of $K^n$, where $e_i$ has $1$ in the $i^{\operatorname{th}}$ coordinate and $0$ elsewhere. It suffices to find a sequence of reflections defined over $K$ whose composition maps $v_i$ to $e_i$ for $i=1,\ldots,r$, since then the inverse sequence maps $e_1,\ldots,e_n$ to an orthonormal basis extending $v_1,\ldots,v_r$. In fact, it suffices to find such a sequence mapping just $v_1$ to $e_1$, since after that we are reduced to an $(n-1)$-dimensional problem in $e_1^\perp$, and can use recursion.

Case 1: $q(v_1-e_1) \ne 0$, where $q$ is the quadratic form. Then reflection in the hyperplane $(v_1-e_1)^\perp$ maps $v_1$ to $e_1$.

Case 2: $q(v_1+e_1) \ne 0$. Then reflection in $(v_1+e_1)^\perp$ maps $v_1$ to $-e_1$, so follow this with reflection in the coordinate hyperplane $e_1^\perp$.

Case 3: $q(v_1-e_1)=q(v_1+e_1)=0$. Summing yields $0=2q(v_1)+2q(e_1)=2+2=4$, a contradiction, so this case does not actually arise.

Any orthonormal set extends to an orthonormal basis, over any field of characteristic not $2$.

Proof: Let $q_n$ be the quadratic form $x_1^2+\cdots+x_n^2$. Translated into the language of quadratic forms, your question is: Given a quadratic form $q$ such that the orthogonal sum $q \perp q_m$ is equivalent to $q_n$ for some $m \le n$, must $q$ be equivalent to $q_{n-m}$? The answer is yes, by the Witt cancellation theorem.

Any orthonormal set extends to an orthonormal basis, over any field of characteristic not $2$. This is a special case of Witt's theorem.