= Solution
{c}
Let $o=(0,0)$ be the common point of the two <circles>. For sufficiently small $\varepsilon>0$, the <open> <neighbourhood>
$$
W=X\cap B_\varepsilon(o)
$$
has the following local structure: $W\setminus\{o\}$ consists of four disjoint <open arcs>, an upper and a lower branch from each <circle>. These are its four <connected components>, each <open> and <closed> relative to $W\setminus\{o\}$.
Suppose a <coordinate chart> of some <dimension of a manifold> $n$ existed at $o$. The case $n=0$ would make $o$ an <isolated point>, which it is not. For $n\geq1$, shrink the chart inside $W$ so that its <image> is an <open ball> centred at the <image> of $o$. Its <domain of a function> $V$ is an <open> <neighbourhood> of $o$, so it meets all four arms of $W$. Intersecting those arms with $V\setminus\{o\}$ gives four disjoint nonempty <clopen sets> partitioning $V\setminus\{o\}$; hence there are at least four <connected components>.
But a <punctured ball> in $\mathbb R$ has exactly two <connected components>, and a <punctured ball> in $\mathbb R^n$ for $n\geq2$ is <path-connected>. A <homeomorphism> preserves <connected components>, giving a contradiction in every <dimension of a manifold>. Thus \b[$X$ is not a <topological manifold>]. The argument uses arbitrarily small <coordinate charts>; the fact that the whole $X\setminus\{o\}$ has two pieces alone would not be a local obstruction.
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