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 $m+n$:
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 $\|\cdot\|$:
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 $m\le x<m+1$; then the integer nearest to $x$ is either $m$ or $m+1$, unless $x$ is halfway between them, in which case it doesn’t matter which we choose, the distance from $x$ is $\frac12$. But then clearly $m+n\le x+n<m+n+1$, and the distances from $x+n$ to $m+n$ and to $m+n+1$ are identical to the distances from $x$ to $m$ and to $m+1$: we’ve simply translated the picture $n$ units. Thus, either $x\le m+\frac12$, and $$\|x+n\|=(x+n)-(m+n)=x-m=\|x\|\;,$$ or $x>m+\frac12$, and
(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 $m\le x<m+1$. Observe that $-(m+1)<-x\le-m$, so $-m$ and $-(m+1)$ are the two integers nearest $-x$. Moreover, the distance from $-x$ to $-m$ is the same as the distance from $x$ to $m$, and the distance from $-x$ to $-(m+1)$ is the same as the distance from $x$ to $m+1$. (I’ll leave the algebraic verification of that to you.) Can you now finish the argument to show that $\|-x\|=\|x\|$?