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I want to prove the following proposition:

Let $A$ be a set, $B$ be a subset of $A$, and $f: Ato B$ be a injective map, then $f(A) = f(A-B) + f(B).$

Could you check my proof below?

Assume $f(A-B)cap f(B) neq emptyset$. For $xin f(A-B)cap f(B)$ there exist $ain A$ which satisfies $f(a)=x$ and $bin A-B$ which satisfies $f(b)=x$. However, this contradicts the original assumption that $f$ is injective: $forall a, b in A, f(a)=f(b)Rightarrow a=b$, thus the above is impossible. Thus, $f(A-B)cap f(B)=emptyset$ and hence $f(A)=f(A-B)+f(B)$.

Other than the typo I've pointed out in a comment, your proof is fine.

Note, however, that you don't need that $B subseteq A$. The following is true:

Lemma. Let $f colon A to B$ be injective, $C subseteq D subseteq A$ Then
$$f[D] = f[D setminus C] mathbin{dot{cup}} f[C].$$

(The proof is virtually the same as the one you've given.)

Also note that injectivity is necessary:

Lemma. If $f colon A to B$ is not injective, then there is some $C subseteq A$ such that
$$f[A setminus C] cap f[C] neq emptyset.$$

I'll leave the easy proof to you as an exercise. (Hint: You can choose $C$ to be a singleton.)

I assume that $+$ denotes disjoint union. Your idea is good, but you lose yourself in explanations.

The proof that $f(A)=f(A-B)cup f(B)$ is easy and doesn't require injectivity.

Now the proof the sets are disjoint. More generally,

if $X$ and $Y$ are disjoint sets and $f(X)cap f(Y)neemptyset$, then $f$ is not injective.

Indeed, if $zin f(X)cap f(Y)$, then $z=f(x)=f(y)$, with $xin X$ and $yin Y$. Since $X$ and $Y$ are disjoint, $xne y$.