I spent a while stuck on this, so now I really want to know if I found a correct solution:
Prove: All $n$-dimensional real inner product spaces are isomorphic to the $n$-dimensional Euclidean space. [Edit: by “isomorphic inner product spaces” (as opposed to just vector spaces) I mean that there exists an isomorphism which preserves the inner product.]
Proof:
Let $V$ be an $n$-dimensional real vector space with inner product $H$. Let $\mathbb{R}^n$ be equipped with some inner product $K$. Using Gram-Schmidt, we can construct orthonormal bases $\beta = \{b_1, \ldots, b_n\}$ for $V$ and $\beta' = \{d_1, \ldots, d_n\}$ for $\mathbb{R}^n$ under their respective inner products. There exists exactly one linear transformation $\phi: V \rightarrow \mathbb{R}^n$ satisfying $\phi(b_i) = d_i$ for $i = 1, \ldots, n$. Since $\phi$ maps a basis for $V$ to a basis for $\mathbb{R}^n$, it's image must be $n$-dimensional. That is, $\phi$ is an isomorphism. Let $x \in \mathbb{R}^n$ have unique scalars $x_1, \ldots, x_n$ such that $x = \sum_{i=1}^n x_i b_i$. Observe that
$$\phi(x) = \sum_{i=1}^n x_i \phi(b_i) = \sum_{i=1}^n x_i d_i.$$
Now we can prove that $\phi$ preserves the inner product: Let $x,y$ be vectors in $V$. Then there exist unique scalars $x_i$ and $y_i$ such that $x = \sum_{i=1}^n x_i b_i$ and $y = \sum_{i=1}^n y_i b_i$. Furthermore,
\begin{align*}
H(x,y) &= \sum_{i,j}^n x_i y_j H(b_i, b_j) = \sum_{i=1}^n x_i y_i H(b_i, b_i) \\
&= \sum_{i=1}^n x_iy_i = \sum_{i=1}^n x_iy_i K(d_i, d_i) \\
&= \sum_{i,j}^n x_i y_j K(d_i, d_j) = K\left( \sum_{i=1}^n x_id_i, \sum_{j=1}^n y_j d_j \right) \\
&= K\left(\phi(x), \phi(y)\right).
\end{align*}
Best Answer
One point is that if "isomorphism" just means "as vector spaces", then the inner product plays no role.
But if "isomorphism" is meant to include not only vector-space isomorphism, but also that the inner product is preserved, then your argument is exactly right.
In more elementary linear algebra situations, often people think purely of vector-space isomorphisms, without the extra structure of an inner product, so a knee-jerk reaction would be that "isomorphism" does not refer to inner products. But it can refer to inner products, as well. "Isomorphism" depends hugely on the context.