Let $A,B$ be two subsets of a set $X$, and let $f:X\rightarrow Y$ be a function. Show that
(a) $f(A\cap B)\subseteq f(A)\cap f(B)$,
(b) $f(A)\backslash f(B) \subseteq f(A\backslash B)$,
(c) $f(A\cup B) = f(A)\cup f(B)$.
MY ATTEMPTS
(a) Suppose that $y\in f(A\cap B)$. Then $y = f(x)$ for some $x\in A\cap B$. Consequently, $y = f(x)$ for some $x\in A$ and $y = f(x)$ for some $x\in B$, which means that $y\in f(A)$ and $y\in f(B)$. Hence we conclude that $y\in f(A)\cap f(B)$.
(b) Let us suppose that $y\in f(A)\backslash f(B)$. Thus we conclude that $y\in f(A)$ and $y\not\in f(B)$. Therefore $y = f(x)$ for some $x\in A$ and $x\not\in B$, which is the same as to state that $y = f(x)$ for some $x\in A\backslash B$. Hence $y\in f(A\backslash B)$, from whence we conclude that $f(A)\backslash f(B)\subseteq f(A\backslash B)$.
(c) Let us prove the inclusion $\subseteq$ first. Indeed, if $y\in f(A\cup B)$, $y = f(x)$ for some $x\in A\cup B$, which means that $x\in A$ or $x\in B$. If $x\in A$, then $y = f(x)$ for some $x\in A$, from whence we infer that $y\in f(A)$. If $x\in B$, then $y = f(x)$ for some $x\in B$, from whence we conclude that $y\in f(B)$. Since $x$ belongs to $A$ or $B$, we conclude that $y$ belongs to $f(A)$ or $f(B)$. That is to say that $f(A\cup B)\subseteq f(A)\cup f(B)$.
Let us now prove the inclusion $\supseteq$. Suppose that $y\in f(A)\cup f(B)$. Then $y\in f(A)$ or $y\in f(B)$. If $y\in f(A)$, there exists an $x\in A\subseteq A\cup B$ such that $y = f(x)$. Similarly, if $y\in f(B)$, then exists an $x\in B\subseteq A\cup B$ such that $y = f(x)$. In both cases, there is an $x\in A\cup B$ such that $y = f(x)$. Consequently, $y\in f(A\cup B)$, which implies that $f(A)\cup f(B)\subseteq f(A\cup B)$, and we are done.
If someone could check if I am reasoning rightly, I would appreciate.