= Solution
{c}
Use positive indices $n\geq1$ and write $p=(0,-1)\in S^1$. Set
$$
\boxed{A=\mathbb N\times\{p\},\qquad f(n,(u,v))=\left(\frac un,\frac{v+1}{n}\right).}
$$
For fixed $n$, this <function> is a <homeomorphism> from the <unit circle> to $C_n$: it scales its <radius> to $1/n$ and translates its centre to $(0,1/n)$. Thus it takes values in the <Hawaiian earring> $X$ and is <surjective> onto $X$. It is <continuous> on each <open> slice $\{n\}\times S^1$; since these slices form an <open cover> of $\mathbb N\times S^1$, $f$ is <continuous> on the whole <product topology>.
Distinct <circles> $C_n,C_m$ meet only at $o=(0,0)$. To see this directly, their equations are
$$
x^2+y^2=\frac{2y}{n},\qquad x^2+y^2=\frac{2y}{m}.
$$
For $m\neq n$, subtraction gives $y=0$, then $x=0$. Each slice contains exactly one <preimage> of $o$, namely $(n,p)$. Therefore the only nonsingleton <fibre> of $f$ is $A$.
Let $q:\mathbb N\times S^1\to(\mathbb N\times S^1)/A$ be the <quotient map> collapsing $A$. The <universal property of the quotient topology> supplies a unique <continuous map> $\bar f$ with $f=\bar f\circ q$. The <fibre> calculation proves <injectivity>, and <surjectivity> follows from that of $f$. Hence \b[$\bar f$ is the required <continuous bijection>].
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