Formalising path connectedness 
Let $L := \{(x,0) \in \mathbb{R}^2 : x \geq 0 \}$.  Prove that $\mathbb{R}^2 \setminus L$ is path connected.
Let $P := \{(x,y) \in \mathbb{R}^2 : y = x^2 \}$.  Prove that $\mathbb{R}^2 \setminus L$ is path connected.

I can visualise this pictorially but I'm struggling to find a rigorous proof, especially for number 1. For 1 we can draw as a straight line for points $p,q > 0$ or $p,q < 0$. When $p$ and $q$ have different signs then we can draw a line to the origin and to $q$.
 A: $\newcommand{\r}{\mathbb R^2 \setminus L}$
Remember the following principle : If there is a path between $a$ and $b$, and a path between $b$ and $c$ , then there is a path between $a$ and $c$(all paths contained within the set). This is done using the usual concatenation operator.
For example, in the first case, it can be seen  visually that every point $p$ can be connected by a path to the point $(-1,0)$. For example, let $p \in \r$ then $p = (a,b)$ where either $a < 0$ or $ b \neq 0$ (or both).
For example, if $a < 0$ then one may take the map $t \to (at + (-1)(1-t) , tb) = (t(a+1) - 1,tb)$ which is the straight line joining $p$ and $(-1,0)$, where we note that $t(a+1) - 1 < 0$ for all $t$ so the line lies entirely in $\r$.
If $b \neq 0$, then the above straight line again avoids $L$ since at $t = 0$ it is $(-1,0) \notin L$ and for $t \neq 0$ we have $tb \neq 0$ so the path avoids $L$. 
Finally, to concatenate $f_1: [0,1] \to \r$ and $f_2 : [0,1] \to \r$ such that $f_1(1) = f_2(0)$, we get $f_3 : [0,1] \to \r$ via:
$$
f_3(t) = \begin{cases}
f_1(2t) \quad \quad \ \ \  t \leq \frac 12 \\
f_2(2t-1) \quad t \geq \frac 12
\end{cases} 
$$ 
One may check that all maps above are continuous, concluding that $\r$ is path connected.

For the second problem, we are given $P$. Upon drawing $P$ it is seen visually that $P$ is path connected but divides $\mathbb R^2$ into two components, so the complement is not path connected. This may be proved easily as follows.
Indeed, given $(x,x^2), (y,y^2) \in P$ , it is easily seen that $t \to (tx + (1-t)y , (tx + (1-t)y)^2)$ is a path between these points which lies entirely in $P$, and is continuous.
For the complement not being connected, let $(a,b)$ and $(c,d)$ be in the complement of $P$. We will show there is no path lying in the complement connecting these points if $a^2 < b$ and $c^2 > d$. Indeed, let $f : [0,1] \to \mathbb R^2$ be any path between these points. Consider the continuous map $F : \mathbb R^2 \to \mathbb R$ given by $F(x,y) = y-x^2$. Note that $F \circ  f : [0,1] \to \mathbb R$ is continuous and has $F \circ f(0) > 0$ and $F \circ f(1) < 0$ therefore there is a point $t \in [0,1]$ with $F \circ f(t) = 0$ . One easily sees therefore that $f(t)$ lies on $P$. Thus, no path exists lying entirely in the complement. 
Read up the Jordan curve theorem , for example.
