= Solution
Let $a,b\in\mathbb R^2\setminus T$. If $a=b$, use the constant <path>. Otherwise consider the countably many lines through $a$ and a point of $T$, and through $b$ and a point of $T$. Choose a line $L$ whose direction is not parallel to any of these lines; such a direction exists because there are uncountably many directions but only a <countable set> of forbidden ones. Every forbidden line meets $L$ in at most one point. Choose $c\in L$ outside those countably many intersections and outside $\{a,b\}$.
Neither segment $[a,c]$ nor $[c,b]$ meets $T$: a forbidden point on either segment would put $c$ on the corresponding forbidden line. The piecewise linear <path>
$$
\gamma(t)=\begin{cases}a+2t(c-a),&0\le t\le1/2,\\c+(2t-1)(b-c),&1/2\le t\le1\end{cases}
$$
lies in the complement. This proves that the <plane minus a countable set is path connected>.
If a <homeomorphism> $h:\mathbb R^2\to\mathbb R$ existed, its restriction after removing a point $a$ would give a <homeomorphism> between $\mathbb R^2\setminus\{a\}$ and $\mathbb R\setminus\{h(a)\}$. The first is a <path-connected space> and therefore <connected>; the second consists of two separated intervals. Since <homeomorphisms> preserve <connectedness>, $\boxed{\mathbb R^2\not\cong\mathbb R}$.
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