For $f: \mathbb{R}\to\mathbb{R}: f(x) = x\sin\left(\frac{1}{x}\right)$, show that $f$ is continuous at all $x\neq 0$. I was advised to complete the proof like so:
\begin{align*} \text{We will show $f$ is continuous at all $c \in \mathbb{R}\setminus\{0\}$.}\\ \forall \epsilon>0 \exists \delta > 0 \mid \forall x: \left|x-c\right|<\delta \Longrightarrow \left|f(x)-f(c)\right|<\epsilon\\ \Longrightarrow \left|x\sin\left(\frac{1}{x}\right)-c\sin\left(\frac{1}{c}\right)\right| < \epsilon\\ \iff \left|x\sin\left(\frac{1}{x}\right)-c\sin\left(\frac{1}{x}\right)+c\sin\left(\frac{1}{x}\right)-c\sin\left(\frac{1}{c}\right)\right|<\epsilon\\ \iff \left|\sin\left(\frac{1}{x}\right)\left(x-c\right)+c\left(\sin\left(\frac{1}{x}\right)-\sin\left(\frac{1}{c}\right)\right)\right|<\epsilon\\ \text{By the Triangle Inequality, we have}\\ \iff \leq \left|\sin\left(\frac{1}{x}\right)\left(x-c\right)\right|+ \left|c\left(\sin\left(\frac{1}{x}\right)-\sin\left(\frac{1}{c}\right)\right)\right|<\epsilon\\ \iff \left|\sin\left(\frac{1}{x}\right)\right|\left|x-c\right| + \left|c\right|\left|\sin\left(\frac{1}{x}\right)-\sin\left(\frac{1}{c}\right)\right|<\epsilon\\ \text{We know that $\sin{x}$ is bounded above by $1$, and $\left|x-c\right|$ is bounded above by $\delta$. Hence we have}\\ \iff < \delta + \left|c\right| < \epsilon\\ \text{Pick $\delta:=1$. Then we have}\\ \Longrightarrow 1 + \left|c\right| < \epsilon\\ \Longrightarrow 0< \epsilon - \left|c\right| - 1 \end{align*} Hence, for $\delta:=\min\left\{1, \left|\epsilon-|c|-1\right|\right\}$, we have $|x-c|<\delta \Longrightarrow |f(x)-f(c)|<\epsilon$. Because $\epsilon>0$ was arbitrarily chosen, it follows that $f$ is continuous at all $c \neq 0$.
Would this proof work? Can I assume that $\left|\epsilon-|c|-1\right|$ can be turned into an absolute value? Somehow this proof does not sit right with me.