Use the Cauchy-Schwarz inequality to prove $||x||_1 \leq \sqrt{n}||x||_2$ If $x,y \in \mathbb{R}^n$, then the Cauchy-Schwarz Inequality tell us that
$$|x^Ty| \leq ||x||_2||y||_2,$$
where
\begin{align*}
\|x\|_2 &= \sqrt{\sum_{i=1}^nx_i^2},\\ 
\|x\|_1 &= \sum_{i=1}^n|x_i|.
\end{align*}
To show that 
$$||x||_1 \leq \sqrt{n}||x||_2,$$
we let $x = (|x_1|, |x_2|, \cdots, |x_n|)$ and $y = (1,1, \cdots, 1)$. 
Then
$$|x^Ty| \leq ||x||_2||x||_2 \Rightarrow |\sum_{i=1}^n|x_i|| \leq \sqrt{\sum_{i=1}^n|x_i|^2}\sqrt{\sum_{i=1}^n1},$$
but for each $i = 1,2, \ldots,n$, we have
\begin{align*}
|x_i| &>0,\\ 
|x_i|^2 &= x_i^2,
\end{align*}
hence it follows that 
$$|\sum_{i=1}^n|x_i|| = \sum_{i=1}^n|x_i| \text{ and } \sqrt{\sum_{i=1}^n|x_i|^2} = \sqrt{\sum_{i=1}^nx_i^2}.$$
Therefore,
$$|\sum_{i=1}^n|x_i|| \leq \sqrt{\sum_{i=1}^n|x_i|^2}\sqrt{\sum_{i=1}^n1} \Rightarrow ||x||_1 \leq \sqrt{n}||x||_2,$$
as required. Is this approach correct?

Yes, this proof is correct. It's interesting that this inequality is actually not strict - consider $x = \left(1, 1, \dots, 1\right)^T$. And, btw, it's a particular case of the theorem stating all norms are equivalent in a finite-dimensional space.
