$f(A) \subseteq f(B)$ whenever $A \subseteq B$ I am looking for feedback if this would constitute a proof, I feel like it is more a explanation than a rigorous proof:
Please provide feedback where it is wrong/inadequate and how it should be done to become rigorous.
Prove:
$f(A) \subseteq f(B)$ whenever $A \subseteq B$
By def 1: $A \subseteq B: \Leftrightarrow((x \in A)\Rightarrow(x \in B))$
By def 2: $f(A)=\{y \in Y \mid y=f(x)$ whenever $x \in A\}$
But when $x \in A$ if follows that $x \in B$.
and since $f(B)=\{y \in Y \mid y=f(x)$ whenever $x \in B\}$
if follows that when $y \in f(A)$ it implies that $y \in f(B)$.
 A: The pieces of the proof are all there, but they're not in order. You're not introducing your variables before you use them, making it unclear what they mean.
First of all, the statement you're proving is incomplete, because it never says what $f$ is, or whether it can be applied to elements of $A$ or $B$. A better statement of what you want to prove would be

Let $f$ be a function $X \to Y$. If $A,B \subseteq X$ with $A \subseteq B$, then $f(A) \subseteq f(B)$.

Your goal is to prove that $f(A) \subseteq f(B)$. The definition of this is that for all $y \in f(A)$, we also have $y \in f(B)$. (Your definitions are missing the "for all", which is important.) The standard structure of such a proof should be:

Let $y$ be an arbitrary element of $f(A)$. Then [...], so $y \in f(B)$. Therefore $f(A) \subseteq f(B)$.

Now it is time to use the definition of $f(A)$. Since $f(A) = \{y \in Y : \exists x \in X, y = f(x)\}$, or more briefly $f(A) = \{f(x) : x \in A\}$, we can now expand on the middle of our proof:

Then, by definition of $f(A)$, we know that there is some $x \in A$ such that $y = f(x)$.

Remember our assumption: that $A \subseteq B$. The definition of this is: every element of $A$ is also an element of $B$. This assumption is useless unless we have an element of $A$ to apply it to. But now we do! That's the $x \in A$ mentioned above. So now we can say:

Since $A \subseteq B$, we also have $x \in B$.

Now we can check that $y \in f(B)$ by checking that the definition holds: that there is some element of $B$ that $f$ sends to $y$. Which element? The element is $x$, of course, because we now know that $x$ is an element of $B$.

Therefore, since $y = f(x)$ and $x \in B$, we have $y \in f(B)$.

Putting these all together:

Let $y$ be an arbitrary element of $f(A)$. Then, by definition of $f(A)$, we know that there is some $x \in A$ such that $y = f(x)$.
Since $A \subseteq B$, we also have $x \in B$. Therefore, since $y = f(x)$ and $x \in B$, we have $y \in f(B)$.
We've shown that an arbitrary element of $f(A)$ is also an element of $f(B)$; therefore $f(A) \subseteq f(B)$.

Note that before we introduce any new variables, we define what they are! This is always important.
Also, before every conclusion we make, we summarize the key details we use to make it. Some of these details can be omitted once you're more comfortable with writing proofs, but you should always include them in the early stages, because you should always be able to include them.
A: You have the right idea.  You can write it a little more cleanly.
Choose $y \in f(A)$.  Then by the definition of $f(A)$, there is some $x \in A$ such that $y = f(x)$.  But $A \subseteq B$, so by the definition of $\subseteq$ we have $x \in A \Rightarrow x \in B$.  Thus, $x \in B$ and so $y = f(x)$ for some $x \in B$. By the definition of $f(B)$, since for some $x \in B$ we have $y=f(x)$, we know that $y \in f(B)$.
Thus, if $y \in f(A)$, we can prove that also $y \in f(B)$.  Again by the definition of $\subseteq$, this means $f(A) \subseteq f(B)$.
