Is the Hausdorff condition redundant here?

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.