Regarding Hatcher's Algebraic Topology Lemma 2.34 In the proof of the Lemma, we first show that $H_k(X^n, X^{n - 1}) = 0$ for $k \neq n$ and free abelian for $k = n$, then use that to show $H_k(X^n) = 0$ for $k > n$. My question is would it also be valid to do the following (assuming that $X$ is finite-dimensional): 
Take $X = X^m$ for some $m$. Then $H_k(X) = H_k(X^m) = H_k(X^{m - 1} \coprod_\alpha e_\alpha^m)$, where $e_\alpha^m$ is an open $m$-disk. But the homology of a disjoint union is the direct sum of homologies, so we find that is equal to $H_k(X^{n - 1} \oplus H_k(e_1^m) \oplus \cdots \oplus H_k(e_n^m)$. But since $e_i^m$ is contractible, $H_k(e_i^m) = 0$, so we find $H_k(X) = H_k(X^i)$ for $i \leq m$, as desired. 
Of course, this argument does not show that $H_k(X^n, X^{n - 1}) = 0$ for $k \neq n$ and free abelian for $k = n$, but I was wondering if this is a valid alternate proof. 
Edit: Wanted to add that $X^n$ is the $n$-skeleton of a CW-complex $X$.
 A: You are right, the homology of the disjoint union of spaces $X_i$ is the direct sum of the homologies of the $X_i$. The topology on the disjoint union $D = \bigsqcup X_i$ is given by declaring $U \subset D$ to be open iff all intersections $U \cap X_i$ are open in $X_i$. Hence all $X_i$ are open in $D$. Morever, they are also closed in $D$ since $D \setminus X_i$ is the union of the open sets $X_j$ for $j \ne i$.
Unfortunalety $X^m = X^{m - 1} \cup \bigcup _\alpha e_\alpha^m$ is not a disjoint union in this sense. All pieces are disjoint, but no $e^m_\alpha$ is closed in $X^m$. In fact, there are attaching maps $\phi^m_\alpha : S^{m-1} \to X^{m-1}$ for all $e^m_\alpha$. Then $X^m$ is the quotient space obtained from $X^{m-1} \sqcup \bigsqcup_\alpha D^m_\alpha$ with $D^m_\alpha = D^m$ by identifying $x \in S^{m-1}_\alpha = S^{m-1}$ with $\phi^m_\alpha(x) \in X^{m-1}$. If  $p$ denotes the quotient map, we have $e^m_\alpha = p(\mathring D^m_\alpha)$. But now $\overline{e^m_\alpha } = \overline{p(\mathring D^m_\alpha }) \supset p(\overline{\mathring D^m_\alpha }) = p(D^m_\alpha) \supset p(S^{m-1}_\alpha) = \phi^m_\alpha(S^{m-1})$, thus $\overline{e^m_\alpha }$ intersects $X^{m-1}$. This shows that $e_\alpha^m$ is not closed in $X^m$.
