The Remmert's proper mapping theorem states
Remmert's proper mapping theorem: Let $f:X\to Y$ be a proper holomorphic map of complex spaces and $A\subset X$ a closed analytic subset. Then $f(A)$ is analytic in $Y$.
As mentioned in the answer of this question, Remmert's proper mapping theorem is a consequence of Grauert's direct image theorem, which states:
Grauert's direct image theorem: Let $f:X\to Y$ be a proper holomorphic map of complex spaces and $\mathcal{F}$ a coherent sheaf on $X$. Then all sheaves $R^qf_*(\mathcal{F})$ are coherent.
By Oka's coherence theorem, the structure sheaf $\mathcal{O}_A$ is a coherent $\mathcal{O}_X$-module. Consequently, $f_*(\mathcal{O}_A)$ is coherent as $\mathcal{O}_Y$-module, whose support is $f(A)$, showing $f(A)$ is analytic.
I'd like to find an example when the analyticity fails without the assumption of properness. Note it is not enough to find an example where Grauert's theorem fails, because without the properness assumption, $supp(f_*(\mathcal{F}))$ can be larger than $f(A)$, and a support of a non-coherent sheaf can still be analytic (Consider the embedding $\mathbb C^*\hookrightarrow \mathbb C$ and push forward of structure sheaf).
Therefore, it seems we should directly work on a counterexample of Remmert's. Consider the closure of the graph $z\mapsto \exp(\frac1z)$ inside $\mathbb C\times \mathbb C$. This space is not analytic because a general horizontal hyperplane section of this space is not finitely generated (a consequence of Great Picard's theorem).
Now, let $Z$ denote the space stated above, and $Y=\mathbb C\times \mathbb C$. My question is that is it possible to find a pair of analytic spaces $A\subset X$, and holomorphic (non-proper) mapping $f:X\to Y$, such that $Z=f(A)$?