What is the relationship between $f(A)$ and $f(B)$ if $A \subseteq B$ but $A\neq B$? Suppose $X$ and $Y$ are sets, that $f : X \to Y$ is a function from $X$ to $Y$ and that $A$ and $B$ are subsets of $X$. Given a subset $C \subseteq X$ let $f(C) = \{y ∈ Y : y = f(c) \text{ for some } c ∈ C\} \subseteq Y$.
I have proved that if $A \subseteq B$ then $f(A) \subseteq f(B)$. 
Proof: let $y \in f(A)$. Then there exists $a \in A \subseteq B$ such that $f(a) = y$. Since $a \in B \Rightarrow f(a) \in f(b) \Rightarrow f(A) \subseteq f(B)$.
Using this as a guide, I would assume $f(A)$ is can still be $\subseteq$ of $f(B)$ because $A \subseteq B$ can either coincide or not coincide but I'm not too sure.
 A: Without any hypothesis on the function $f \colon X \to Y$, you can only say that, given $A, B \subseteq X$, if $A \subseteq B$ then $f(A) \subseteq f(B)$, as you have already proven.
You can't say that $A \subsetneq B$ implies $f(A) \subsetneq f(B)$ (where $C \subsetneq D$ means that $C$ is a proper subset of $D$, i.e. $C \subseteq D$ and $C \neq D$, for all sets $C$ and $D$); in other words, you can have that $A \subsetneq B$ but $f(A) = f(B)$. This is an example: let $f \colon \{0,1\} \to \{2\}$ be the function defined by $f(0) = 2$ and $f(1) = 2$. Then, $\{0\} \subsetneq \{0,1\}$ but $f(\{0\}) = \{2\} = f(\{0,1\})$.
Anyway, if you suppose that moreover the function $f \colon X \to Y$ is injective (i.e. $x_1 \neq x_2$ implies $f(x_1) \neq f(x_2)$), then you can prove also that $A \subsetneq B$ implies $f(A) \subsetneq f(B)$. Proof: We already know that $f(A) \subseteq f(B)$, so we have just to prove that $f(A) \neq f(B)$. Since $A \subsetneq B$, there exists a $b \in B \smallsetminus A$; then $f(b) \in f(B)$ but $f(b) \notin f(A)$, otherwise there would be an $a \in A$ (and hence $a \neq b$ since $b \notin A$) such that $f(a) = f(b)$, which is impossible because $f$ is injective. Therefore, $f(A) \neq f(B)$.
