Topology of CW-complex and attaching map I think I must have a fundamental misconception in place right now in my mind.
When defining a CW-complex, we use inductively continuous maps from $f_{\partial \sigma} :S^n \to K^{(n)}$. We then define a natural map $f_{\sigma}: D^{n+1} \to K^{(n+1)}$ which proceeds with the construction. For this, we take the disjoint union $\bigsqcup\limits_{\sigma} D_{\sigma}^{n}$ over the cells we are attaching and define
$$K^{(n+1)} =  \left(\bigsqcup\limits_{\sigma} D_{\sigma}^{n} \right) \cup_f K^{(n)}  .$$
Now, why do we need $f_{\partial \sigma}$ to be continuous?  The definition of the topology on $K^{(n+1)}$ does not require continuity of $f_{\partial \sigma}$. We have a disjoint union, and a quotient map, which defines the quotient topology. Also, the $f_{\sigma}$ seem to be continuous, even if $f_{\partial \sigma}$ is not. This would follow from the fact that $f_{\sigma}$ is equal to $\pi \circ i$, where $\pi$ is the quotient map and $i$ is the inclusion map on the disjoint union, and those maps are continuous. $f_{\sigma}$ being continuous implies $f_{\sigma}|_{S^{n}}$ being continuous, since it is just the restriction. But such restriction is commonly considered as the attaching map $f_{\partial \sigma}$ itself, which (by what I said above) need not be continuous for such definitions to make sense. The problem at hand, it seems, is that the induced topology on $K^{(n)}$ as a subspace is not the same as the topology of $K^{(n)}$ itself, and maybe this is why we require the attaching maps to be continuous (in order for restriction to preserve the expected topologies), but I'm not sure.
My question, summing up the issues, is:
What is the relevance of the attaching map being continuous? If it is not, what kind of issues arise?
 A: 
The problem at hand, it seems, is that the induced topology on $K^{(n)}$ as a subspace is not the same as the topology of $K^{(n)}$ itself, and maybe this is why we require the attaching maps to be continuous (in order for restriction to preserve the expected topologies), but I'm not sure.

Yes, this is exactly right.  If the attaching maps were not continuous, then when you formed $K^{(n+1)}$, you would be modifying the topology of $K^{(n)}$ to make them continuous.  You can get all sorts of terrible spaces if you don't require this.
For instance, suppose you decided to attach a 2-cell to $S^1$ via some horrible discontinuous map $f:S^1\to S^1$ which is surjective when restricted to every nontrivial interval of the domain.  In the resulting space, this map $f$ becomes continuous.  But the only ordinary open subsets $U\subseteq S^1$ such that $f^{-1}(U)$ is open are $U=\emptyset$ and $U=S^1$ (since the inverse image of any point under $f$ is dense!).  This means that in the quotient space, the subspace $S^1$ now has the indiscrete topology!  In particular, the quotient is not Hausdorff.
An enormous amount of the utility of CW-complexes comes from the fact that they can be understood inductively using their skeleta.  For instance, cellular (co)homology (and more generally the Atiyah-Hirzebruch spectral sequence) comes from considering a CW-complex as being filtered by its skeleta.  But this only works when you give the skeleta their subspace topologies, and those subspace topologies are much more difficult to understand if they do not coincide with the topologies of the skeleta as CW-complexes themselves.  (For instance, using the subspace topologies, it would no longer necessarily be the case that the quotient $K^{(n)}/K^{(n-1)}$ is a wedge of $n$-spheres--in the example above, $K^{(1)}$ is a circle but its subspace topology is indiscrete!)
