Let $\left( e_n \right)$ be an orthonormal sequence in an inner product space $X$. Then for every $x \in X$, we have $$ \sum_{n=1}^\infty \left\vert \langle x, e_n \rangle \right\vert^2 \ \leq \ \Vert x \Vert^2, $$ where $\Vert \cdot \Vert$ denotes the norm induced by the inner product. This is called the Bessel inequality.
Now how to derive the following inequality, called the Schwarz inequality, from the Bessel inequality?
For any elements $x, y \in X$, $$ \vert \langle x, y \rangle \vert \ \leq \ \Vert x \Vert \cdot \Vert y \Vert. $$
In the finite-dimensional case, this is not so hard as in this case the orthonormal sequence will have only finitely many (distinct) terms, say, $\{e_1, \ldots, e_n\}$, and the last set will then become a basis of $X$.
So every $x, y \in X$ have the unique representations $$x = \sum_{i=1}^n \langle x, e_i \rangle e_i$$ and $$y = \sum_{i=1}^n \langle y, e_i \rangle e_i,$$ respectively, as a linear combination of the $e_i$s.
Then we would have $$\Vert x \Vert^2 = \langle x, x \rangle = \sum_{i=1}^n \left\vert \langle x, e_i \rangle \right\vert^2,$$ $$\Vert y \Vert^2 = \langle y, y \rangle = \sum_{i=1}^n \left\vert \langle y, e_i \rangle \right\vert^2,$$ and $$\langle x, y \rangle = \sum_{i=1}^n \left( \langle x, e_i \rangle \cdot \overline{\langle y, e_i \rangle} \right);$$ so that $$ \begin{align*} \left\vert \langle x, y \rangle \right\vert &= \left\vert \sum_{i=1}^n \left( \langle x, e_i \rangle \cdot \overline{\langle y, e_i \rangle } \right) \right\vert \\ &\leq \sum_{i=1}^n \left( \left\vert \langle x, e_i \rangle \right\vert \cdot \left\vert \langle y, e_i \rangle \right\vert \right) \\ &\leq \sqrt{\left( \sum_{i=1}^n \left\vert \langle x, e_i \rangle \right\vert \right)} \cdot \sqrt{\left( \sum_{i=1}^n \left\vert \langle y, e_i \rangle \right\vert \right) } \\ &= \Vert x \Vert \cdot \Vert y \Vert. \end{align*} $$
Is this reasoning correct?
And if so, then how to tackle the case when $X$ is infinite-dimensional?
What you are doing is actually using Schwarz inequality to prove Schawarz inequality.
In the last step you need to use the Holder's Inequality as follows \begin{align} |\langle x,y\rangle|& = \biggl{|}\sum_\limits{n=1}^{\infty}\langle x,e_1\rangle \overline{\langle y,e_i\rangle}\biggr{|}\\ &\leq \sum_\limits{n=1}^{\infty}|\langle x,e_1\rangle|| \overline{\langle y,e_i\rangle}|\\ &\leq \sqrt{\sum_\limits{n=1}^{\infty}|\langle x,e_1\rangle|^2} \sqrt{\sum_\limits{n=1}^{\infty}|\langle y,e_1\rangle|^2} && \text{ (Holder's Inequality)}\\ &\leq \|x\|\|y\| && \text{(Bessel's Inequality)}. \end{align}
$\textbf{EDIT:}$ Sorry for this not helping answer. An easy way out is as follows:
Let $x\in X$ and let $y\neq 0$ be given. Then just take $e=\frac{y}{\|y\|}$. This $\{e\}$ only makes an orthonormal system. So, by Bessel's Inequality it will follow that $|\langle x,e\rangle|^2 \leq \|x\|^2$. Now just multiply both sides by $\|y\|$ to get $|\langle x,y\rangle|^2\leq \|x\|\|y\|$.