 Solution

Let $W=(\mathbb{N}\times S^1)/A$, and choose in each circle the point opposite $o$:In the subspace topology from the Euclidean plane, the sequence $z_n$ converges to $o$, which does not belong to $B$. Thus $B$ is not closed in $X$.

On the other hand, let $D={\bar{f}}^{-1}(B)\subseteq W$. With the quotient map $q$ from part (i),This is closed in $\mathbb{N}\times S^1$: its complement is the open set $\mathbb{N}\times (S^1\setminus \{(0,1)\})$. The definition of the quotient topology, applied to complements, says that $D$ is closed in $W$. A homeomorphism would carry this closed set to the closed set $B$, a contradiction. Therefore $\bar{f}$ is not a homeomorphism.

The obstruction occurs at the common point. The open set $W\setminus D$ contains the collapsed point, but its image $X\setminus B$ contains no neighbourhood of $o$, since $z_n\to o$. The wedge sum topology allows a separate small neighbourhood on each circle; a neighbourhood of $o$ in the Hawaiian earring must contain all sufficiently small whole circles.