Yes, you do need to go back to the definition of $\|x\|$. To prove that $\|\alpha+\beta\|\le\|\alpha\|+\|\beta\|$ for all $\alpha,\beta\in\Bbb R$, for instance, you have to show that the integer closest to $\alpha+\beta$ cannot be larger than the sum of the integers closest to $\alpha$ and to $\beta$. Here’s one way in which you might approach the task.
Let $m$ and $n$ be the integers closest to $\alpha$ and $\beta$, respectively. (If either $\alpha$ or $\beta$ is halfway between two consecutive integers, you can use either of those integers.) Then $m+n$ is at least a reasonable candidate for the integer nearest to $\alpha+\beta$, so let’s have a look at the distance from $\alpha+\beta$ to $m+n$:
$$\begin{align*}
|\alpha+\beta-(m+n)|&=|(\alpha-m)+(\beta-n)|\\
&\le|\alpha-m|+|\beta-n|\\
&=\|\alpha\|+\|\beta\|\;.
\end{align*}$$
We wanted to show that $\|\alpha+\beta\|\le\|\alpha|+\|\beta\|$, so if we can show that $$\|\alpha+\beta\|\le|\alpha+\beta-(m+n)|\;,$$ we’re done. But this is immediate from the definition of $\|\cdot\|$:
$$\begin{align*}
\|\alpha+\beta\|&=\min\{|\alpha+\beta-k:k\in\Bbb Z\}\\
&\le|\alpha+\beta-(m+n)|\;,
\end{align*}$$
since the minimum of a set is always less than or equal to each member of the set.
Showing that $\|x\|=\|x+n\|$ for each $n\in\Bbb Z$ is easier, because it’s easier to come by the right intuition. Let $m=\lfloor x\rfloor$, so that $m\le xm+\frac12$, and
$$\|x+n\|=(m+n+1)-(x+n)=(m+1)-x=\|x\|\;.$$
(There are more efficient ways to say this; I’m concentrating on how you might actually discover correct arguments.)
Finally, what about $\|-x\|=\|x\|$? Once again let $m=\lfloor x\rfloor$, so that $m\le x