We know that a finite basis of the finite-dimensional space $\mathbb{R}^n$ is
$$ \{(1, 0, 0, 0,\ldots,0),\:(0, 1, 0, 0, 0,\ldots,0),\:(0, 0, 1, 0, 0, 0,\ldots, 0),\:\ldots,\:(0, 0, \ldots, 0, 0, 0, 1)\} $$
What about the infinite-dimensional space $\mathbb{R}^{\infty}$? I want to know that how is the proof of the infinity basis of $\mathbb{R}^{\infty}$?
One of the most frequently considered infinite-dimensional spaces is $\ell^2(\mathbb R)$, which is the set of all infinite sequences $\mathbf x = (x_1,x_2,x_3,\ldots)$ of real numbers for which $x_1^2+x_2^2+x_3^2+\cdots<\infty$. One sort of "basis" used for that space is $\{(1,0,0,0,\ldots), (0,1,0,0,\ldots), (0,0,1,0,\ldots),\ldots)$. This is an example of a Hilbert space.
But that is not a "basis" by one of the most frequently used definitions of "basis": some elements of $\ell^2(\mathbb R)$ cannot be written as a finite linear combination of members of this "basis". One speaks of certain infinite "linear combinations", but then one must be careful about which kind of convergence is used. The kind most often used in thinking about this space involves the norm $\|\mathbf x\|=\sqrt{x_1^2+x_2^2+x_3^2+\cdots}$. One says that $\mathbf x_n$ converges to $\mathbf x$ if $\|\mathbf x_n-\mathbf x\|\to0$.
This is of course not the only such "basis".
If you haven't thought about infinite-dimensional spaces before, it may be rash to assume they're just like finite-dimensional spaces, and your question shows evidence that you're doing that.