Your idea is generally correct.
Using the axiom of choice, $|X|=|X|+|X|$, so there is a bijection between $X$ and $X\times\{0,1\}$. Clearly the latter can be partitioned into two uncountable sets, $X\times\{0\}$ and $X\times\{1\}$.
Therefore $X$ can be partitioned to two uncountable disjoint sets.
Indeed you need the axiom of choice to even have that every infinite set can be written as a disjoint union of two infinite sets, let alone uncountable ones.
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
The statement is false as noted by @Hanul Jeon in comments.
Consider the following uncountable collection of uncountable disjoint subsets of $\mathbb{R}^2$:
$$\mathcal{U}=\big\{\{x\}\times\mathbb{R}\ \big|\ x\in\mathbb{R}\big\}$$
Then consider any bijection $f:\mathbb{R}^2\to\mathbb{R}$ and note that
$$f(\mathcal{U})=\big\{f(U)\ \big|\ U\in\mathcal{U}\big\}$$
is an uncountable collection of pairwise disjoint uncountable subsets of $\mathbb{R}$.