I was trying to prove the implication from right to left. I say that a space $X$ is compactly generated if $X$ is T2 and the collection $\mathcal{K}$ of the compact sets of $X$ is a fundamental cover (I say that a cover $\mathcal{K}$ of $X$ is fundamental if given a set $A$, $A\mbox{ is open}\Leftrightarrow A\cap K\mbox{ is open }\forall K\in \mathcal{K}$).
For the other one this should work (correct me if I'm wrong). Let $C$ be a closed set in $Y$, then I $f(C)\mbox{ is closed} \Leftrightarrow f(C)\cap K \mbox{ is closed in } K \; \forall K\in \mathcal{K}$, but $f(C)\cap K=f(C\cap f^{-1}(K))$ where $C$ is closed for hypothesis and $f^{-1}(K)$ is closed and compact (because I have a continuous proper map). But $C\cap f^{-1}(K) \subseteq f^{-1}(K)$ and so $C\cap f^{-1}(K)$ is compact because it is a closed subspace of a compact subspace. But then $f(C\cap f^{-1}(K))=f(C)\cap K$ is compact in a T2 space and thus closed.
How could I proceed for proving the other implication? Thanks in advance
The left to right direction is OK. But it's easily transformed into a direct proof:
Suppose $f: Y \rightarrow X$ is proper. Let $C$ be closed in $Y$. Then for a compact subset $K$ of $X$ we have: $f[C] \cap K = f[f^{-1}[K] \cap C]$ which is a compact set as the continuous image of a compact set, and thus closed in $K$. As $X$ is compactly generated, $f[C]$ is closed in $X$ and $f$ is closed.
For the other implication: suppose that $C$ is such that $C \cap K$ is closed in $K$ for all compact subsets $K$ of $X$. Let $i:C \rightarrow X$ be the embedding $i(x) = x$. which is always continuous and now proper as $i^{-1}[K] = K \cap C$ which is compact as a closed subset of $K$. Now $i$ is closed, so in particular $i[C] = C$ is closed in $X$.