$\ell^2$ itself is a simple example of an infinite-dimensional separable Hilbert space.
The orthonormal basis is just a maximal orthonormal set -- so a Hamel basis of a dense subspace which is, in addition, orthonormal. It can certainly be countable.
Even in finite dimensions you can easily change the inner product. Let $\{ e_{n} \}$ be an orthonormal basis of $\mathbb{C}^{N}$ and define the new inner product
$$
(x,y)_{\mbox{new}}=\sum_{n=1}^{N}\lambda_{n}(x,e_{n})(e_{n},y).
$$
where $\lambda_{n} > 0$ for all $n$.
All the norms are equivalent on $\mathbb{C}^{N}$. This can be written as
$$
(x,y)_{\mbox{new}}=(Ax,y)_{\mbox{old}}
$$
where $A$ is a positive definite selfadjoint matrix. In finite dimensions, this describes every possible inner-product. The inner products are in one-to-one correspondence with positive definite matrices. Because selfadjoint $A$ can be diagonalized by an orthonormal basis of eigenvectors, $(Ax,y) = \sum_{n}\lambda_{n}x_{n}y_{n}^{\star}$ always looks like a weighted inner product when viewed with respect to a correctly chosen orthonormal basis.
If $X$ is an infinite dimensional linear space on which two topologically equivalent Hilbert inner products are defined, say $(\cdot,\cdot)$ and $(\cdot,\cdot)_{1}$, then the same thing happens. There exists a unique positive bounded selfadjoint $A$ such that
$$
(x,y)_{1}=(Ax,y),\;\;\; x,y\in X.
$$
But it also goes the other way: $(x,y)=(Bx,y)_{1}$ where $B$ is positive. You end up with $(x,y)=(Bx,y)_{1}=(ABx,y)$ which gives $AB=I$. Similarly $BA=I$. The existence of such $A$ and $B$ comes from the Lax-Milgram Theorem, which is proved using the Riesz Representation Theorem for bounded linear functionals on a Hilbert Space.
But there are bounded positive linear operators $A$ on a Hilbert space $X$ which are not positive definite. Such an $A$ gives rise to $(x,y)_{1}=(Ax,y)$ with $\|x\|_{1} \le C\|x\|$ for a constant $C$, but the reverse inequality need not hold, which can lead to an incomplete $X$ under $\|\cdot\|_{1}$. For example, let $X=L^{2}[0,1]$ with the usual inner product. Define a new inner product by $(f,g)_{1}=\int_{0}^{1}xf(x)g(x)\,dx$. This is achieved as $(Af,g)$ where $Af=xf(x)$. This space is not complete because $1/\sqrt{x}$ is in the completion of $L^{2}$ under the norm $\|\cdot\|_{1}$. The completion of $(X,\|\cdot\|_{1})$ consists of $\frac{1}{\sqrt{x}}L^{2}[0,1]$. However, if you instead define $\|f\|_{1}^{2}=\int_{0}^{1}(x+\epsilon)|f(x)|^{2}\,dx$ for some $\epsilon > 0$, then you end up with an equivalent norm on $X$.
Best Answer
If we are given an inner product on a finite dimensional space, there will always be a change of basis that turns the given inner product into the Euclidean one.
By definition the product $(\cdot,\cdot)$ can be represented as a symmetric positive-definite matrix $A$ such that $(x,y) = x^T A y$. Now by the spectral theorem (which you probably learnt in Linear Algebra) the matrix $A$ has an orthogonal basis of eigenvectors. In other words there is an orthogonal matrix $M$ such that $J = M^TAM$ is diagonal. That means $(x,y) = x^T M^TJ My = (Mx)^TJ(My)$.
Exercise: Use positive-definiteness to prove $J$ has only positive diagonal entries $\lambda_1,\lambda_2,\ldots,\lambda_n$.
It follows there is a further diagonal matrix $L$ such that $J=LIL = L^TIL$. That gives $$(x,y) = (Mx)^TL^TIL(My)= (LMx)^T(LMy)$$ Put another way, for each $x,y$ we have
$$\big(\,(LM)^{-1}x,(LM)^{-1}y\, \big) = x^T y$$
That means the given product is the Euclidean product with respect to the basis with transition matrix $LM$.