Can the Hilbert space from Gaussian kernel admit a basis with uncountable vectors? According to Mercer's theorem, for $\mathbf{x},\mathbf{y}\in\mathbb{R}^d$, the symmetric positive-definite kernel $\exp\left(-\|\mathbf{x}-\mathbf{y}\|^2/(2\sigma^2)\right)$ admits an orthogonal basis $\{\phi_i(\cdot)\}_i$ with countable vectors, such that 
$$
\exp\left(-\frac{\|\mathbf{x}-\mathbf{y}\|^2}{2\sigma^2}\right)=\sum_{i=1}^\infty \lambda_i\phi_i(\mathbf{x})\phi_i(\mathbf{y})
$$
But I can "find" another basis with uncountably many eigenfunctions as follows.
First, the Fourier transform of a Gaussian function is Gaussian. So if changing the variable $\mathbf{s}$ in the transform 
$$
\exp\left(-\frac{\|\mathbf{s}\|^2}{2\sigma^2}\right)=C\int_{\mathbb{R}^d}\exp\left(-\frac{\sigma^2\|\mathbf{t}\|^2}{2}\right)\exp\left(-i\,\mathbf{s}^\mathrm{T} \mathbf{t}\right)\,\mathrm{d}\mathbf{t}
$$
to $\mathbf{x}-\mathbf{y}$, we can have
$$
\exp\left(-\frac{\|\mathbf{x-y}\|^2}{2\sigma^2}\right)=C\int_{\mathbb{R}^d}\exp\left(-\frac{\sigma^2\|\mathbf{t}\|^2}{2}\right)\exp\left(-i\,\mathbf{x}^\mathrm{T} \mathbf{t}\right)\exp\left(i\,\mathbf{y}^\mathrm{T} \mathbf{t}\right)\,\mathrm{d}\mathbf{t}
$$
which means $\exp\left(-\|\mathbf{x-y}\|^2/(2\sigma^2)\right)$ can be expressed as an inner product of $\exp\left(-i\,\mathbf{x}^\mathrm{T} \mathbf{t}\right)$ and $\exp\left(-i\,\mathbf{y}^\mathrm{T} \mathbf{t}\right)$, with weight $C\exp\left(-\sigma^2\|\mathbf{t}\|^2/2\right)$. So $\phi(\mathbf{x})=(\exp(-i\,\mathbf{x}^\mathrm{T}\mathbf{t}))_\mathbf{t}$, an infinite-dimensional vector indexed by $\mathbf{t}\in\mathbb{R}^d$, can be seen as a basis for the Hilbert space from kernel $\exp\left(-\|\mathbf{x}-\mathbf{y}\|^2/(2\sigma^2)\right)$ (since the inverse Fourier transform of $0$ is $0$, the vectors are linear-independent). 
So is there anything wrong in the above "proof"? It seems to me that a vector space with countable basis vectors cannot hold uncountable linear-independent vectors.
 A: The expression $$\tag{1}\exp\left(-\frac{\|\mathbf{x}-\mathbf{y}\|^2}{2\sigma^2}\right)=\sum_{i=1}^\infty \lambda_i\phi_i(\mathbf{x})\phi_i(\mathbf{y})$$ is an equality between functions. The functions $\phi_i$ are orthonormal. You never bother to mention and/or consider with respect to what inner product. The context of Mercer's theorem is that you are considering kernels of Hilbert Schmidt operators on $L^2(\mathbb R^d)$. So the $\phi_i$ are "orthonormal" in the sense that 
$$
\delta_{k,j}=\langle \phi_j,\phi_k\rangle:=\int_{\mathbb R^d}\phi_j(\mathbf t)\phi_j(\mathbf t)\,d\mathbf t.
$$
In your second approach you want to consider a function as a vector, indexed by its variable. Nothing wrong with that. But you say that the kernel " can be expressed as an inner product": note that $(1)$ is not an inner product with the inner product that is considered in the Hilbert space where your Hilbert-Schmidt operator acts. Also, you say that  $\{\exp(-i\mathbf x^T\mathbf t)\}_{\mathbf t}$ is a basis: of what? With what inner product? For a fixed $\mathbf x$?
In the end, I'm just guessing, but I think that your confusion arises from seen $(1)$ as an inner product (which you could say it is, in $\ell^2$, for each fixed $\mathbf x,\mathbf y$, but it is not relevant to the context), when it is not really related with the underlying inner product of your Hilbert space. 
