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
Option $2.$ is not correct. Consider the set $A=\big\{(x,\sin\big(\frac{\pi}{x}\big):0<x\leq 1\big\}$ in $X=\Bbb R^2$. Then $\overline A=A\cup \big(\{0\}\times[-1,1]\big)$ which is not path connected.
To prove this, note that, $A$ is path connected as it is graph of a continuous function on the path connected set $(0,1]$.
Next, if possible $\overline A$ is path connected. Then there is a path $\gamma :[0,1]\to A$ with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,0)$. So $\pi_1\gamma,\pi_2\gamma$ are continuous, $\pi_1,\pi_2:\Bbb R^2\to \Bbb R$ are projections on $1$-st and $2$-nd coordinates. But, $\pi_1\gamma$ takes all values of the form $\frac{1}{n},n\in\Bbb N$ by intermediate value property of $\pi_1\gamma$, as $\pi_1\gamma(0)=0,\pi_1\gamma(1)=1$ . So $\pi_2\gamma$ assumes each values $\pm 1$ in every nbd of $0\in [0,1]$. So there is no nbd $[0,\delta)$ of $0$ in $[0,1]$ which maps to $\big(-\frac{1}{2},\frac{1}{2}\big)$ under the map $\pi_2\gamma$, that is $\pi_2\gamma$ is discontinuous.