So $X=\{x_1\dots x_n\}$ is a complex inner product space,
and $Y\subseteq X$ is a complete orthonormal set.
Hence $(y_i,y_j)=0$ for $y_i\ne y_j$, and $||y||=1$ for all $y\in Y$
We can also write $Y=\{y_1\dots y_m\}, m\le n$,
and we know that $Y$ is the largest such set in $X$
Now how to show that $Y$ is a basis for $X$?
Presumably you'd start by showing each $x\in X$ can be expressed in the form $\displaystyle{\sum_{i=1}^m a_i y_i}$, where $a_i\in\mathbb{C}\,\forall\,i=1\dots m$
But I don't know how you'd go about doing that, and I don't know if that would be sufficient. Any help appreciated.
I will start by saying that your notation for $X$ is misleading, that is writing it as a sequence. In future, just write $X$ is an inner product space of dimension $n$. Now what you want to prove seems a little redundant as the definition of a complete orthonormal set is a generating set in the algebraic sense. That is, if $Y$ is a complete orthonormal set of $X$, then by definition for every $x\in X$ there exists $c_{1},\ldots,c_{n}\in\mathbb{C}$ such that, \begin{align} x=\sum_{i=1}^{n}c_{i}y_{i}. \end{align} In this sense, there is nothing to prove, since by definition $Y$ is a basis of $X$.