Proof: Let any element

of
be given; by definition, each element of
may be approximated by such elements. Let
. Then by definition of an orthonormal basis, we find
for
and
for
and then
resp.
such that
and
.
Then note that by the triangle inequality,
.
Now fix
. Then by the triangle inequality,

In total, we obtain that

(assuming that the given sum approximates
well enough) which is arbitrarily small, so that the span of tensors of the form
is dense in
.
Now we claim that the basis is orthonormal. Indeed, suppose that
. Then
.
Similarly, the above expression evaluates to
when
and
. Hence,
does constitute an orthonormal basis of
.