Prove that $f(\{x\})=\{f(x)\}$ for all $x\in X$
Suppose $\gamma\in f(\{x\})$. Then there exists $\alpha\in\{x\}$ such that $f(\alpha)=\gamma$; namely, $\alpha=x_i$ for some $x_i\in X$. Thus, $\gamma=f(\alpha)=f(x_i)$; that is, $\gamma\in\{f(x)\}$.
Suppose $\gamma\in\{f(x)\}$. Then $\gamma=f(x_i)$ for some $x_i\in X$. That is, $\gamma\in f(\{x\})$.
Hence, $f(\{x\})=\{f(x)\}$.
Does this look correct? Any suggestions on how to improve it perhaps or have I gone wrong anywhere?
On
You need to prove that given a specific x, f({x})={f(x)}. What you proved is that given x, if γ∈f({x} then there is some xi∈X such that γ∈{f(xi)}. That is also true but does not prove the needed equality, since you want to have xi = x .
Suppose γ∈f({x}). Then f(x)=γ. Thus, γ∈{f(x)}.
Suppose γ∈{f(x)}. Then γ=f(x) Thus, γ∈f({x}).
So we see that f({x})={f(x)}
On
In this context think of $x$ as a fixed element of $X$.
It can all be done shorter (see the answers) but on purpose I stick to your line of reasoning.
If $\gamma\in f(\{x\})$ then $\gamma=f(y)$ for some $y\in\{x\}$.
For $y$ there is only one choice: $y=x$.
So eventually we find $\gamma=f(x)$ or equivalently $\gamma\in\{f(x)\}$.
Proved is now that: $$f(\{x\})\subseteq \{f(x)\}\tag1$$
If $\gamma\in\{f(x)\}$ then automatically $\gamma=f(x)$ or equivalently $\gamma\in\{f(y)\mid y\in\{x\}\}=f(\{x\})$.
Proved is now that: $$\{f(x)\}\subseteq f(\{x\})\tag2$$
Combining $(1)$ and $(2)$ we find:$$f(\{x\})=\{f(x)\}\tag3$$
This has been proved for an arbitrary fixed $x\in X$ allowing the conclusion that $(3)$ is true for every $x\in X$
Let's just unwrap the definitions. If $f\colon X \to Y$ and we choose any $x \in X$, observe that: \begin{align*} f(\{x\}) &= \{f(\alpha) \in Y \mid \alpha \in \{x\}\} \\ &= \{f(\alpha) \in Y \mid \alpha = x\} \\ &= \{f(x)\} \\ \end{align*}