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Before we begin, let's define some key terms: - Inner product space: A vector space equipped with an inner product, a function that assigns a scalar to two vectors. - Hilbert basis: A set of vectors whose linear combinations approximate any given vector in the space arbitrarily closely. - Zorn's lemma: In a nonempty partially ordered set (poset) in which every chain (totally ordered subset) has an upper bound, there exists at least one maximal element. Now we will prove that any nontrivial inner product space has a Hilbert basis using Zorn's lemma.

Let's consider the set of all linearly independent subsets of our inner product space "S". We need to prove that "S" is a nonempty partially ordered set (poset). For this, we need to define an order relation on the elements of "S". We can use the ⊆ (subset) order relation on the linearly independent sets. According to this relation, given two sets A and B, A ⊆ B if every element of A is also an element of B. To show that "S" is nonempty, note that any vector subspace contains the empty set ∅. The empty set is linearly independent because it satisfies the property that no linear combination of vectors in the set can equal to the zero vector. Therefore, "S" is nonempty.

Let "C" be a chain in "S" (a totally ordered set of linearly independent subsets of the inner product space). We want to show that "C" has an upper bound. Consider the union of all sets in the chain "C", denoted as U(C). We claim U(C) is an upper bound for "C" (meaning that every set in "C" is a subset of U(C)). Since the elements in "C" are linearly independent sets and "C" is a chain, any two sets A, B ∈ C must satisfy A ⊆ B or B ⊆ A. Thus, the union U(C) includes all elements from every set in "C". Now we need to prove that U(C) is linearly independent. Suppose there exists a linearly dependent vector in U(C), then it can be written as a linear combination of other vectors in U(C). However, this contradicts the fact that all vectors in the sets of "C" are linearly independent. Thus, U(C) is linearly independent. Since the chain "C" has an upper bound in "S", we can now apply Zorn's lemma, which states that there must be at least one maximal element in "S".

Let M be a maximal element in "S". We will show that M constitutes a Hilbert basis for our inner product space. If M spans the entire inner product space, then M is indeed a Hilbert basis. Suppose there exists a vector "v" in the inner product space that is orthogonal to every vector in M, where v ≠ 0. Note that adding "v" to M creates a larger linearly independent set, since "v" cannot be a linear combination of vectors in M. However, this contradicts the maximality of M in "S". Therefore, there are no such orthogonal vectors "v" outside M. This means that any vector in the inner product space can be approached arbitrarily closely by a linear combination of vectors in M. Consequently, M constitutes a Hilbert basis for the inner product space.