Is the Hausdorff condition redundant here? This is a question in Algebraic Topology by Hatcher, Chapter 0:

21) If $X$ is a connected Hausdorff space that is a union of a finite number of 2 spheres,
  any two of which intersect in at most one point, show that $X$ is homotopy equivalent
  to a wedge sum of $S^1$’s and $S^2$’s.

Is it necessary that Hatcher requires $X$ to be Hausdorff here?  Is it possible to have a space $X$ as a finite disjoint union of spaces homeomorphic to $S^2$ where we identify at most one point of each pair and $X$ not be Hausdorff.  It seems pretty obvious to me that this is not possible but maybe I'm overlooking something.

I know this is an old question, but here is an example which only uses two $2$-spheres. Let $X$ be a union of two $2$-spheres with a point in common, and let $i,j:S^2\to X$ be the two inclusions of $S^2$ into $X$. Fix a point $p\in S^2$ to be the basepoint and let $i(p)=j(p)$ be the point of intersection of the two spheres.
We topologize $X$ by declaring the sets $i(U)\cup j(U)$ to be open for all $U$ open in $S^2$. This differs from the topology on the wedge in that all open sets have to be "symmetric."
 This is a topology. One can check the axioms directly. The empty set and whole space are clearly open. An intersection of symmetric open sets is symmetric, and an arbitrary union of symmetric open sets is symmetric. Furthermore, restricting this topology to either sphere gives the standard topology on the sphere.
It is however, highly non Hausdorff. The points $i(x)$ and $j(x)$ cannot be separated for $x\neq p$.
Edit: The example I constructed can be thought of as a quotient of $S^2\times\{a,b\}$, where $\{a,b\}$ is given the coarse topology. One collapses one of the fibers $p\times\{a,b\}$ to a point. From this description it becomes obvious that in addition to the two embeddings of $S^2$ corresponding to $i$ and $j$ above, there are in fact uncountably many embeddings of $S^2$ into $X$, given by arbitrarily picking $a$ or $b$ for each point in $S^2$. This is not particularly relevant for the problem, but I thought it was interesting.
