Proposition Any path-connected CW-complex is homotopy equivalent to a CW-complex with precisely one 0-cell.
Proof (Sketch) Let $X$ be a path-connected CW-complex, so $sk_1(X)$ is a connected graph. Let $\Gamma\subset sk_1(X)$ be a spanning tree. Since $\Gamma$ is contractible, the map $X\to X/\Gamma$ is a homotopy equivalence.
Question What kind of quotient space is $X/\Gamma$? How should I visualize it?
Let $X$ be a CW-complex and $A$ a subcomplex, that means $A$ is a union of cells $\{e_\alpha^n\mid\alpha\in J \}$ such that $\overline{e_\alpha^n}$ is also contained in $A$. Then $A$ is closed and the quotient map $q:X\to X/A$ is a closed surjection. Such maps preserve the $T_4$-property of $X$. In particular, $X/A$ is Hausdorff. We can then apply the following proposition which translates between the intrinsic and the constructive definition of a CW complex:
As the new characteristic maps we take $\Psi_\beta=q\circ\Phi_\beta$ for each cell $e_\beta^n$ which is not in $A$, and a single map $D^0\to\{A\}$ as the new $0$-cell the set $A$ was shrunk to. These maps clearly satisfy condition (i). It is also easy to see that they satisfy (ii) since the image of $\partial D_\alpha^n$ can only be in fewer cells after collapsing $A$. Condition (iii) says that $X$ has the final topology of all $\Phi_\alpha$, but since $q$ is a quotient map, also $X/A$ will have the final topology of all $\Psi_\beta$.
So this gives us a natural cell complex structure on the quotient space.