Any set containing an unbounded interval is path-connected. The intuitiv idea is that you can walk to infinity and jump from there to any point you like.
E.g. if it contains $(y_0,\infty)$ for some $y_0\in \mathbb{R}_{>0}$. Let $x_0\in \mathbb{R}$. If $x_0\geq y_0$, then they are clearly connected by a continuous path ($\gamma:[0,1]\rightarrow X, \gamma(t)= x_0+ t(y_0-x_0$)). On the other hand, if $x_0< y_0$, then we have the path
$$ \gamma: [0,1] \rightarrow X, \gamma(t)=\begin{cases} \frac{y_0}{t},& t\neq 0, \\ x_0,& t=0. \end{cases} $$
Let me show that is continuous in the case $x_0\geq 0$ (the case $x_0<0$ is similar). Note that any open nbhd $U$ of $x_0$ in $X$ can be written as
$$ U= V \setminus \{x_0\} \cup (-\infty, -m) \cup (\{x_0\} \cup (n, \infty)) $$
where $m,n\in \mathbb{R}_{>0}$ and $V\subseteq \mathbb{R}$ open and bounded. Then we have
$$ \gamma^{-1}(U) = \gamma^{-1}(V\setminus \{x_0\}) \cup \gamma^{-1}((-\infty, -m)) \cup \gamma^{-1}(\{x_0\} \cup (n,\infty)).$$
We show that all of those sets are open. First we note $\gamma^{-1}((-\infty, -m))=\emptyset$, which is open in $[0,1]$. Next we have
$$ \gamma^{-1}(\{x_0\} \cup (n,\infty)) = \begin{cases} [0, \frac{y_0}{n}),& n>y_0, \\ [0,1],& n\leq y_0. \end{cases} $$
In both cases the sets are open in $[0,1]$.
Finally, as $V$ is bounded, there exists $R>0$ such that $V\subseteq (-R,R)$. Then we define
$$ \tau : [0,1] \rightarrow \mathbb{R},\tau(t):= \min \{ R, \gamma(t) \}. $$
As $\tau$ is continuous and $\tau^{-1}(V\setminus \{x_0\} ) = \gamma^{-1}(V\setminus \{x_0\})$, we get that also $\gamma^{-1}(V\setminus \{x_0\})$ is open in $[0,1]$.
Right now I do not have time to work it out, but I guess that that if a set does not contain an unbounded interval, then it is path-connected in $X$ iff it is path-connected in $\mathbb{R}$. The idea is that the space is first-countable and thus continuity and sequential continuity coinced (see here Sequentially continuous implies continuous). Then we should be able to use that bounded sequences in $X$ converge iff they converge in $\mathbb{R}$. This means you cannot jump unless you are at infinity.
Best Answer
Bizarrely, I think the space $(\mathbb R,\mathcal O)$ is path-connected: If $r \in \mathbb R\backslash \mathbb Q$ and $q\in \mathbb Q$. let $\gamma\colon [0,1]\to \mathbb R$ be given by $\gamma(t)=q$ if $t<1$ and $\gamma(1)=r$. If $U$ is an open subset of $(\mathbb R,\mathcal O)$ then $\gamma^{-1}(U)$ is either
1). $\gamma^{-1}(U) = \emptyset$ if $\{q,r\}\cap U = \emptyset$,
2). $\gamma^{-1}(U) = [0,1)$ if $q\in U$ and $r \notin U$;
3). $\gamma^{-1}(U) = [0,1]$ if $r \in U$, since $x\in U$ implies $\mathbb Q \subseteq U$ because $r \notin \mathbb Q$.
It follows that the path component of $r$ contains all $q \in \mathbb Q$, and since this is true of all $r \in \mathbb R\backslash \mathbb Q$, it follows that $(\mathbb R,\mathbb Q)$ is path connected: for example, if $q_1,q_2 \in \mathbb Q$ then there are paths from both $q_1$ and $q_2$ to $\sqrt{2}$, and hence a path from $q_1$ to $q_2$, and similarly if $r_1,r_2$ are irrational, there are paths from both $r_1$ and $r_2$ to $0$, and hence a path from $r_1$ to $r_1$.