Let $f:X \to Y$ be an unrammified covering map of degree $N$. How do you prove that $\chi(X) =N\chi(y)$. We use the following definition of $\chi$.
We say that $\chi(\{x\})=-1, \chi\left(D(0,1)\right)=1, \chi([0,1])=1$ and $\chi(A\setminus{B}) = \chi(A)- \chi(B).$ This is basically how we define $\chi$. From the last equation, it follows that for $A$ and $B$ disjoint, that $\chi(A\cup B) = \chi(A) +\chi(B)$. I'm also assuming that this characteristic is topological invariant.
Now, for each point in $Y$, it has an open neighbourhood $U$ such that it's inverse image is a disjoint union of open sets, where each set $V_\alpha$ is homeomorphic to $U$. We say that $U$ is evenly covered. The inverse image of $Y$ is X, but you could choose "one pancake out of the stack of pancakes", with $Y$ as image. With this I mean you can cover $Y$ with all the evenly covered subsets, and when we look at the inverse image of this cover, we can select one specific $\alpha$ i.e. $\alpha=1$ in this case, so $X$ would have a subspace, which is covered by all those $V_1$, we call the part of $X$ that is covered $X_1$. I'm not quite sure but, I want to say that $X$ is a disjoint union of those $X_\alpha$, and that $X_\alpha \cong Y$, and therefore $\chi(X) = N \chi(X_\alpha) = N\chi(Y)$.
How do you prove this. And where does stuff go wrong?
if $X,Y$ are CW (you can adapt for simplicial) complexes, then for each cell $e^k \to Y$ there are exactly $n$ lifts to $X$ ***, since the restriction of $p\mid_{p^{-1}(e^k)}$ is a covering map as well (the only one up to homotopy.) Hence, if we take the alternating sum over, say, cellular homology $H_i(X,\mathbb Z_2)$ then this gives the result readily.
***(these exist by the lifting lemma) applied to the induced map on fundamental group $\pi_1(e^k) \to \pi_1(y)$, but $e^k$ is contractible.
regarding your argument: there is no reason to suspect that $X$ is the disjoint union of spaces homeomorphic to $Y$! For example, the covering $S^2 \to \mathbb RP^2$ gives a quick counterexample to that idea. On the other hand, it gives an opportunity to verify the theorem. In particular, taking the usual cellular decomposition of $\mathbb RP^2$, you can check that $\chi(\mathbb RP^2)=1-1+1=1$ while $\chi(S^2)=2$, which is correct since the cover is of degree $2$.