It seems like there are decomposable rays which limit onto themselves.
Let $E=\bigcup_{n\geq0}E_n$ and $O=\bigcup_{n\geq0}O_n$, with $E_n=[2n,2n+1]$ and $O_n=[2n+1,2n+2]$. We want to construct a ray $f:[0,\infty)$ which limits onto itself and such that $f(E),f(O)$ are closed in the ray $X$ and connected.
For $f(E)$ to be connected it will be enough to ensure that $\forall x\in f(E)\;\forall\varepsilon>0\;\exists N$ such that $\forall n>N$, $d(x,f(E_n))<\varepsilon$. Indeed, if that happens any clopen of $f(E)$ which contains $x$ will have to contain $f(E_n)$ for $n$ big enough, so it will contain all $f(E)$.
For $f(E)$ to be closed in $X$ it will be enough to assign a small tubular neighborhood $B_n$ to the set $f(O_n^o)$ (where $O_n^o$ is the interior of $O_n$) such that $f(E)$ doesn't intersect $B_n$.
The construction will be inductive. Set $f([0,2])$ as some arbitrary smooth curve. Now given $n\geq1$, suppose that we have defined $f$ in $[0,2n]$ injectively and we have nhoods $A_i$ of $f(E_i^o)$ and $B_i$ of $f(O_i^o)$, for $i=1,\dots, n-1$, such the sets $A_i$ don't intersect $f(O\cap[0,2n])$ and the $B_i$ don't intersect $f(E\cap[0,2n])$. Suppose also that the $A_i$ are pairwise disjoint and bounded by simple closed curves (and the same for the $B_i$).
This means that the set $R_n=\mathbb{R}^2\setminus (f([0,2n])\cup \cup_{i=1}^{n-1}\overline{B_i})$ is homeomorphic to the annulus $(0,1)\times\mathbb{S}^1$, and its boundary is a non injective curve formed by the $f(E_i)$ and the boundaries of the $B_i$ (similarly for $S_n=\mathbb{R}^2\setminus (f([0,2n])\cup \cup_{i=1}^{n-1}\overline{A_i})$).
We will define $f(E_n)$ as a smooth injective curve in $R_n$ (except for the initial point $f(2n)$). We can define this curve so that it passes at distance $<2^{-n}$ of every point of the set $f(E\cap[0,2n])$, which is contained in the boundary of $R_n$. Similarly, we define $f(O_n)$ as a smooth curve in $S_n\setminus f(E_n)$ (which is homeomorphic to an annulus) which passes at distance $<2^{-n}$ of every point of $f(O\cap[0,2n])$. Then define $A_n,B_n$ to be small tubular neighborhoods of $f(E_n^o)$ and $f(O_n^o)$ such that the conditions above are satisfied.
This could be modified to make the curve nowhere dense: at each step we can enlarge $A_n$ and $B_n$ so that they contain some point $p_n$ which is not in $f([0,2n+2])$. If we take a sequence $p_n$ dense in the plane, the ray won't intersect a nhood of each $p_n$, so it will be nowhere dense. If for some $n$, $p_n$ is already in $\overline{A_i}$ for some $i<n$, we wouldn't need to enlarge $A_n$, and the same for $B_n$ (this is to keep the $A_i$ pairwise disjoint).
Best Answer
There is no such bijection.
A line in the plane is almost the same as a plane through the origin in 3-space (by intersecting with the plane at height 1), except there's one plane through the origin that doesn't give you a line (the z=0 plane). So the space of lines in the plane is homeomorphic to $\mathbb{RP}^2$ minus a point: an open mobius strip! So the question is asking if there is a continuous bijection from the open disk $D$ to the open mobius strip $M$. Invariance of domain implies that a continuous bijection between manifolds of the same dimension is a homeomorphism. $D$ and $M$ are not homeomorphic, so there cannot be a continuous bijection between them.