Show that $f(x)$ is continuous at $x = 0$...

Question

We have $f(x) = $

\begin{cases} x\sin(\frac{1}{x}) & x \neq0 \\ 0 & x = 0 \end{cases}

My proof is as follows:

Let $\epsilon > 0$ be given. Choose $\delta = \epsilon$. $\forall \,0 < |x| < \delta$, we have:

$|f(x) - f(0)| = \Bigl|x\sin(\frac{1}{x}) - 0\Bigr| = \Bigl|x\sin(\frac{1}{x})\Bigr|$

Now, $|\sin\alpha| \leq 1$ for all $\alpha \in \mathbb{R}$, so:

$\Bigl|x\sin(\frac{1}{x})\Bigr| = |x|\,\Bigl|\sin(\frac{1}{x})\Bigr| \leq |x| < \delta = \epsilon$

QED.

Is this proof correct?