I propose that a function $f: X\rightarrow Y$ is injective if and only if for all subsets $A,B$ of $X$, $f(A)\cap f(B)\subset f(A\cap B)$.
Let $f$ be $1-1$ and $A,B\subset X$. Let $y\in f(A)\cap f(B)$. Then there is $x_A\in A$ and $x_B\in B$ such that $y=f(x_A)=f(x_B)$. Since the function is injective, it must be the case that $x_A=x_B=x$ (say). $x\in A\cap B$. So $y=f(x)\in f(A\cap B)$. Therefore $f(A)\cap f(B)\subset f(A\cap B)$.
Suppose now, that for all subsets $A,B$ of $X$ that $f(A)\cap f(B)\subset f(A\cap B)$ holds. Let $x_1,x_2\in X$ such that $f(x_1)=f(x_2)=y$ (say). Let $A:=${$x_1$} and $B:=${$x_2$}. If possible let $x_1≠x_2$. Then $A\cap B=\phi \implies f(A\cap B)=\phi.$ But, $f(A)\cap f(B)=${$y$}, which is nonempty. This contradicts the hypothesis. So $x_1=x_2$. It follows that $f$ is injective.
Is the proof correct? Is there a simpler proof or an easy way to understand/visualize why this is true?