Covering $\mathbb{R}^2$ with uncountably many disjoint non-degenerate line segments Is it possible to cover $\mathbb{R}^2$ with uncountably many disjoint non-degenerate line segments? 

If a formal definition is necessary, let's define a line segment as a set $\{(x, mx+c): x \in [a, b]\}$ for some fixed constants $m, c, a, b \in \mathbb{R}$ with or a set $\{(u, x): x \in [a, b]\}$ for some fixed constants $u, a, b \in \mathbb{R}$. We say a line segment so defined is non-degenerate if $a \neq b$, i.e. the line segment is not a point.

This question was vaguely motivated by the observation that it's possible to cover $\mathbb{R}$ with uncountably many disjoint non-degenerate points. YuvalFilmus points out that the answer is negative in the one-dimensional case.
The answer is YES, you can cover the plane by non-degenerate line segments.

It is clear one can cover


any $1 \times 1$ closed squares (i.e those isometric to $[0,1] \times [0,1]$ ) by line segments of length $1$. 
any $1 \times n$ quarter-open rectangles ( i.e. those isometric to $[0,1) \times [0,n]$ ) by line segments of length $n$.


You then proceed to cover $\mathbb{R}^2$ by 


first placing a $1 \times 1$ closed square in the center.
surround the $1 \times 1$ closed square by four $1 \times 2$ quarter-open rectangles.
The "open" sides of the rectangles will be facing "inwards" and the
five quadrilaterals together form a $3\times 3$ closed square.
surround the $3 \times 3$ closed square by four $1 \times 4$ quarter-open rectangles.
The "open" sides again facing "inwards" and the nine quadrilaterals together form a
$5 \times 5$ closed square.
Just repeat this procedure. If you have a $(2k-1) \times (2k-1)$ closed square, you
surround it by four $1 \times 2k$ rectangles to form a $(2k+1) \times (2k+1)$ closed square.


Following is a picture illustrating the arrangement of the squares/rectangles
of line segments. 


The line segments are represented by color bars. 
The black lines indicate the the boundaries of the squares/rectangles.
Slice $\mathbb R^2$ into uncountably many copies of $\mathbb R^1$.