Continuous and preserves measurability $\implies$ preserves null sets. Let $X$ be a (Lebesgue-)measurable set of $\mathbb{R}^n$ and $f:X \to \mathbb{R}^n$ continuous function that preserves measurability ($A$ meausurable $\implies f(A)$ measurable).
Prove: for all $A \subset X$,$\space$$\lambda(A)=0 \implies \lambda(f(A)) = 0$ .
I'm totally stuck. Initially I made some progress but now I'm at a point where it feels like the statement shouldn't be true at all.
Additionally i'd like to know if the statement works for general topological measure spaces (or even just metric measure spaces). That is: 
Let $X$ be a (Lebesgue-)measurable set of $Y$, Topological (or metric) measure space and $f:X \to Y$ continuous function that preserves measurability etc...
 A: Prove this by contrapositive:
Suppose that $f$ doesn't preserve null sets.  Then there is a set $f(N)$ that does not have zero measure.  That is, $f(N)$ must have a non-measurable subset $A$.
Now, consider $f^{-1}(A) \cap N$, which must be measure zero, by the completeness of Lebesgue measure.
A: For the convenience of others, I'll try present Omnomnomnom's complete idea. 
Suppose $f$ preserves measurability and doesn’t preserve null sets. So there is an $N$ so that $N$ is a null set and $f(N)$ is not. Since $N$ is a null set, $N$ is measurable and so $f(N)$ is measurable with positive measure. This implies that there exists a nonmeasurable subset $A \subset f(N)$.
$A \subset f(N) = f|_{N}(N)$, so $$f(f|_{N}^{-1}(A)) = f|_{N}(f|_{N}^{-1}(A)) = A.$$
Since $f(f|_{N}^{-1}(A)) = A$, $f$ preserves measurability, and $A$ is nonmeasurable, we conclude $f|_{N}^{-1}(A)$ is nonmeasurable.
Since $N \cap f^{-1}(A) \subset N$, and $m(N) = 0$, completeness of the Lebesgue measure implies $N \cap f^{-1}(A)$ is measurable (with measure $0$).
But from 
$$
\begin{align*}
x \in f|_{N}^{-1}(A) &\Leftrightarrow x \in N \mathrm{\ and\ } f|_{N}(x) \in A \\ 
&\Leftrightarrow x \in N \cap f^{-1}(A),
\end{align*}
$$
we find $f|_{N}^{-1}(A) = N \cap f^{-1}(A),$ a contradiction.
Note that we didn't use continuity.
A: I will use the hint from @David Mitra. Knowing that

A set of positive measure contains a non-measurable subset


Any subset of a null-set is a null-set

by the contrapositive of the first and using the second you can state that

Let $A$ be a Lebesgue-measurable set. Every subset $B\subset A$ is measurable, if and only if, $m(A)=0$. Even more, $m(B)=0$.

Going back to proving the statement of the question, our strategy to show that $m(f(A))=0$ will be to show that every subset $D\subset f(A)$ is measurable. If we take pre-image: $f^{-1}(D)\subset A$. But we know that $m(A)=0$, so by the proposition before said, $f^{-1}(D)$ is measurable. Taking the image of the previous set $D=f(f^{-1}(D))$. But as $f$ preserves measurable sets and $f^{-1}(D)$ is, so do $D$.
We never used the continuity of $f$.
