Assume $X$ is a Hilbert space, $\{e_n\}_{ n=1}^{\infty}$ is an orthogonal set of $X$. Let $x=∑α_ne_n$， $y=∑β_ne_n$.
Prove $(\mathrm{x}, \mathrm{y})=\sum_{\mathrm{n}=1}^{\infty} \alpha_{\mathrm{n}} \bar{\beta}_{\mathrm{n}}$, and the series on the right of the equality is absolute convergent.
I use a method： Let $\alpha=\sum_{n=1}^{\infty}\left|\alpha_{n}\right| e_{n} \quad \beta=\sum_{n=1}^{\infty}\left|\beta_{n}\right| e_{n}$ Then take $(α,β)$ and then get the conclusion(for a inner product is a certain value). Is this proof right? I don't think the 'Hilbert' is useful here. Could I just extend this conclusion to a normol inner product space?
(I used "orthonormal" for your $\{e_n\}$ instead of "orthogonal", which adds a few complications)
Repeat with me: a series is not sum. A series is a limit of sums. So to be able to talk about series you need to be able to talk about limits: thus you need a topology. And, equally importantly, for the limits to exist you need your space to be complete. In an arbitrary inner product space, you will likely have Cauchy sequences which are not convergent. That's why we care about Hilbert spaces.
In particular, in a non-complete inner product space you might have (using your notation from above) that $x\in X$ but $\alpha$ doesn't exist.
That said, your method will lead nowhere: there is no useful relation between your $\alpha$ and your $x$. What they want you to do here is the following: first show that $\sum_n\alpha_n\overline{\beta_n}$ converges absolutely. This is due to Cauchy-Schwarz: $$ \sum_n|\alpha_n\overline{\beta_n}|=\sum_n|\alpha_n|\,|{\beta_n}|\leq\left(\sum_n|\alpha_n|^2\right)^{1/2}\left(\sum_n|{\beta_n}|^2\right)^{1/2} $$ and the two series on the right converge because $x$ and $y$ exist. Namely, you know that for any $\varepsilon>0$, there exists $n_0$ such that for any $n_1,n_2>n_0$ $$ \left\|\sum_{n=n_1}^{n_2} \alpha_ne_n\right\|^2<\varepsilon. $$ The left-hand-side above is precisely $\sum_{n=n_1}^{n_2}|\alpha_n|^2$. So you obtain that $\sum_n|\alpha_n|^2<\infty$. Similarly with $y$ and $\beta_n$.
Now using that the inner product (proof below) is continuous (you might have to prove this, too) \begin{align} (x,y)&=\left(\sum_n\alpha_ne_n,\sum_n{\beta_n}e_n\right)=\lim_m\left(\sum_{n=1}^m\alpha_ne_n,\sum_{n=1}^m{\beta_n}e_n\right)\\[0.3cm] &=\lim_m\sum_{k,n=1}^m\alpha_n\overline{\beta_k}(e_n,e_k) =\lim_m\sum_{n=1}^m\alpha_n\overline{\beta_n} =\sum_{n=1}^\infty\alpha_n\overline{\beta_n} \end{align}
Proof that the inner product is continuous: suppose that $z_n\to z$ and $w_n\to w$ in $X$. Then, using Cauchy=Schwarz, $$ (z_n,w_n)-(z,w)=(z_n-z,w_n)+(z,w_n-w)\leq \|z_n-z\|\,\|w_n\|+\|z\|\,\|w_n-w\|\to0 $$ if we show that $\|w_n\|<c$ for a fixed $c$. This follows from $w_n\to w$. For $n$ big enough, $\|w_n-w\|<1$. Then $\|w_n\|=\|w_n-w+w\|\leq\|w_n-w\|+\|w\|<1+\|w\|$.