 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 pathlies 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, $\boxed{{\mathbb{R}}^2\ncong \mathbb{R}}$.