Calculate: $ \lim_{x \to 0 } = x \cdot \sin(\frac{1}{x}) $ Evaluate the limit:
$$ \lim_{x \to 0 } = x \cdot \sin\left(\frac{1}{x}\right) $$
So far I did: 
$$
\lim_{x \to 0 } = x\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}\cdot x}
$$
$$
\lim_{x \to 0 } = 1 \cdot \frac{x}{x}
$$
$$
\lim_{x \to 0 } = 1
$$
Now of course I've looked around and I know I'm wrong, but I couldn't understand why. Can someone please show me how to evaluate this limit correctly? And tell me what I was doing was wrong.

Your proof is incorrect, cause you used incorrect transform, but it has already been stated. I'll describe way to solve it.
$$\lim_{x \to 0}\frac{\sin(\frac{1}{x})}{\frac{1}{x}} \neq 1$$
Hint: Solution is well-known trick. Note $(\forall x \in \mathbb{R})\left(\sin(x) \in[-1;1]\right)$ (obvious) and use squeeze theorem to solve it.

Note simple implication.
$$ \left(\forall h \in \mathbb{R}\right) \left(\sin h \in [-1;1]\right) \Longrightarrow (\forall x,h \in \mathbb{R})(|x \cdot \sin h| \leq |x|)$$
So, true is inequality $|x \cdot \sin \frac{1}{x}| \leq |x|$, therefore (and because module is always non-negative) using squeeze theorem you receive limit.
$$\left(0 \leq\left | \lim_{x \to 0} x\cdot \sin \frac{1}{x} \right | \leq \lim_{x \to 0} \left| x \right| = 0 \right)\Longrightarrow \lim_{x \to 0}x \cdot \sin(x) = 0$$
