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.
 A: Note that

a union of a finite number of 2 spheres, any two of which intersect in at most one point

is not the same as

a space $X$ as a finite disjoint union of spaces homeomorphic to $S^2$ where we identify at most one point of each pair,

the union Hatcher speaks of need not arise as a quotient of a disjoint union (topological sum).
As a set, let $X$ be the union of four spheres of radius $1$ with centres in $(\pm 1, \pm1, 0)$ in $\mathbb{R}^3$. In each of the four points $(1,0,0),\, (-1,0,0),\, (0,1,0),\, (0,-1,0)$ two of the spheres intersect. Now endow that space with a coarser topology than the subspace topology, that is not Hausdorff, but nevertheless induces the standard topology on each of the spheres.
For a point $x$ that is not an intersection point of two neighbouring spheres, let the neighbourhoods of $x$ be the intersections of a neighbourhood of $x$ in $\mathbb{R}^3$ with $X$, a neighbourhood basis is then
$$\mathscr{N}(x) = \left\lbrace B_{\varepsilon}(x) \cap X : \varepsilon > 0 \right\rbrace.$$
For an intersection point $p$, let a neighbourhood basis of $p$ be
$$\mathscr{N}(p) = \left\lbrace \bigl(B_\varepsilon(p) \cup B_{\varepsilon'}(-p)\bigr) \cap X : \varepsilon > 0, \varepsilon' > 0\right\rbrace.$$
Verify that there is a unique topology on $X$ such that the above are neighbourhood bases for the respective points, and that topology is not Hausdorff (every neighbourhood of an intersection point $p$ meets every neighbourhood of $-p$).
Yet $X$ is the union of finitely many (four) two-spheres (subspaces homeomorphic to $S^2$), any two of which intersect in at most one point.
So it is not redundant to require $X$ to be Hausdorff.
A: 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.
