There is a theorem about the finite-dimensional inner product space.
Suppose a finite-dimensional inner product space $V$ with a subspace $W$, then $V=W\bigoplus W^{\bot}$.
And the proof is as follows: Suppose an orthonormal basis of $W$ is $u_i, \cdots, u_m$, then for $\alpha \in V$: $$(\alpha-\sum_{i=1}^{m}a_iu_i,\sum_{j=1}^{m}b_ju_j)=0.$$ $$\Leftrightarrow(\alpha-\sum_{i=1}^{m}a_iu_i,u_j)=0, j=1,2,\cdots, m.$$ $$\Leftrightarrow a_i=(\alpha,u_i).$$ $$\Leftrightarrow \alpha-\sum_{i=1}^{m}(\alpha,u_i)u_i\in W^{\bot}$$ And since every nonzero vector in $W^{\bot}$ is indenpendent with the vector in $W$. So the above decomposition is direct sum decomposition.
But I am confused if this proof itself can be generalized to the infinete-dimensional inner product space. Actually, I was always confused what properties can be generalized into infinite-dimensional space throughout the linear algebra learning. Can you give me some direct or give some reference that about that. Thank you in advance.
An infinite dimensional inner product space with an inner product (and the additional property of completeness) is called a Hilbert space. It turns out that in a Hilbert space, your statement does not necessarily hold true. In particular a subspace $W$ of a Hilbert space $\mathcal H$ will satisfy $\mathcal H = W \oplus W^\perp$ if and only if $W$ is (topologically) closed.
For example, we necessarily have $\mathcal H = W \oplus W^\perp$ when $W$ is a finite dimensional subspace of $\mathcal H$. To be more specific about where the direct generalization of your proof fails, a subspace $W$ that fails to be closed does not have a Schauder basis.
The study of infinite dimensional vector spaces like this one falls under the domain of functional analysis. If you are interested in a relevant reference, you might want to try reading Kreyszig's Introductory Functional Analysis with Applications, which I find to be "beginner friendly" (yet fairly comprehensive) relative to similar texts.