$\epsilon$-$\delta$ proof that $f(x) = x \sin(1/x)$, $x \ne 0$, is continuous I'm doing an exercise that asks me to prove that $f$ is continuous using a $\epsilon$-$\delta$ proof. I have that
$$
  f(x) = \begin{cases}
x\cdot \sin \frac1x,&x\neq 0
\\
0,&x = 0
\end{cases}
$$
I've already managed to show this property for $x=0$. How can I show it for $x \ne 0$, also using a $\epsilon$-$\delta$ proof?
Thank you very much.

Although the answer given by @Nameless is complete in itself, a slight change in the choice of added terms in the triangle inequality helps a lot. A modified version of his answer is as follows.
Consider
\begin{align*}
\big|f(x)-f(a)\big|&=\left|x\sin\frac{1}{x}-a\sin\frac{1}{a}\right|\\
&=\left|x\sin\frac{1}{x}-x\sin\frac{1}{a}+x\sin\frac{1}{a}-a\sin\frac{1}{a}\right|\\
&=\left|x\left(\sin\frac{1}{x}-\sin\frac{1}{a}\right)+\sin\frac{1}{a}(x-a)\right|\\
&\leq|x|\left|\sin\frac{1}{x}-\sin\frac{1}{a}\right|+\left|\sin\frac{1}{a}\right|\big|x-a\big|
\end{align*}
By the trigonometric identity $\sin\alpha-\sin\beta=2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$, we have
\begin{align*}
\left|\sin\frac{1}{x}-\sin\frac{1}{a}\right|&=\left|2\sin\frac{\frac{1}{x}-\frac{1}{a}}{2}\cos\frac{\frac{1}{x}-\frac{1}{a}}{2}\right|=2\left|\sin\frac{a-x}{2ax}\cos\frac{x+a}{2ax}\right|\\
&\leq2\left|\sin\frac{a-x}{2ax}\right|\leq2\left|\frac{x-a}{2ax}\right|=\frac{|x-a|}{|a||x|}
\end{align*}
Therefore,
\begin{align*}
\big|f(x)-f(a)\big|&=|x|\left|\sin\frac{1}{x}-\sin\frac{1}{a}\right|+\left|\sin\frac{1}{a}\right|\big|x-a\big|\\
&\leq\frac{|x-a|}{|a|}+\frac{|x-a|}{|a|}=\frac{2|x-a|}{|a|}
\end{align*}
Thus we see that for any $\varepsilon>0$, we can find $\delta=\frac{|a|\varepsilon}{2}$ such that
$$\big|f(x)-f(a)\big|<\varepsilon\qquad\text{whenever}\qquad|x-a|<\delta$$
showing that $f$ is continuous at all points $a\neq0$
