Proof of $f(\{x\})=\{f(x)\}$ I thought, is this really that simple? Or am I missing a piece?
This is my proof:
$f(\{x\})=\{f(x)\}$.
Look at $f(\{x\})$. By definition, $f(\{x\})=\{f(a)|a \in \{x\}\}$, and therefore $f(\{x\})=\{f(x)\}$.
 A: The definition of a function is a set of ordered pairs, such that if for some $x$ we have $\langle x,y\rangle,\langle x,z\rangle\in f$ then $y=z$.
That is to say, $f$ is a binary relation and for every $x$ there is at most one element such that $\langle x,y\rangle\in f$. We call often say that $y$ is $f(x)$.
Now, what is $f(A)$ when $A\subseteq\operatorname{Dom}(f)$? It is the set $\{f(x)\mid x\in A\}$. Let $A=\{x\}$, we have that $$f(\{x\}) = \Big\lbrace f(a)\mid a\in\{x\}\Big\rbrace$$ 
Since $a\in\{x\}\iff a=x$ we have that $f(\{x\}) = \{f(x)\}$ as wanted.
A: This is basically the same proof as yours, I just rewrote it more formally/at lower level.
To prove that $f(\{x\})=\{f(x)\}$ we need to show that:
$$z\in f(\{x\}) \Leftrightarrow z\in \{f(x)\}.$$
This can be shown as follows:
$z\in f(\{x\})$ $\Leftrightarrow$ $(\exists a\in\{x\}) z=f(a)$ $\Leftrightarrow$ $z=f(x)$ $\Leftrightarrow$ $z\in\{f(x)\}$
(You might want to think for a bit why the equivalences I wrote above are true - in case you want to make a really detailed proof. I think that most of the details are explained nicely in Asaf's answer.)
