Indeed, there are closed and connected subsets of a finite-dimensional Euclidean space which are not arc-wise connected, as it is a continuous image of L.
edit (after Benoît Kloeckner's comment) It seems to me that a complete metric space $X$ which is connected and locally connected is necessarily arc-wise connected. The idea is that given two points $x$ and $y$ in $X$, and $\epsilon>0$, there is a finite sequence $ x=x^\epsilon_0,x^\epsilon_1,\dots ,x^\epsilon_n=y$ such that for $0\le i < n$ the points $x^\epsilon_i$ and $x^\epsilon_{i+1}$ belong to some connected open set $U^\epsilon_i$ of diameter less than $\epsilon$ (the reason is that given $x$, the set of all $y$ which are reachable this way is an open and closed non-empty set). So we can start with $\epsilon = 1$, and iterate the construction within each $U^\epsilon_i$, which is still connected and locally connected, finding new points between $x^\epsilon_i$ and $x^\epsilon_{i+1}$, taking $\epsilon=1,1/2,1/4\dots$. By completeness these dotted lines converge to a suitably parametrized arc joining $x$ and $y$.
edit. The preceding is indeed exactly Whyburn theorem (1931). In particular a closed, connected, locally connected, subset $A$ of the Euclidean space is arc-wise connected. You further ask if it is a continuous image of the real line. If $A$ is bounded then you can even obtain it as a continuous image of the closed unit interval, via a construction à la Peano (incidentally, you can, of course, also choose the endpoints of the arc). More generally, continuous images of the unit closed interval are caracterized by the Hahn-Mazurkiewicz theorem. If $A$ is unbounded, you may write it as a countable union of bounded compact sets, each one image of an arc with domain $[k,k+1]$. These arcs glue together in a continuous function on $\mathbb{R}$ provided you choose the endpoints so that they match.
Finally, there are connected and locally connected subsets of the Euclidean space, in dimension at least 2, which are not countable union of compact sets, hence they are not continuous images of the real line. An example is $A=(\mathbb{R}\times\mathbb{R})\setminus (\mathbb{Q}\times\mathbb{Q})$, thanks to the Baire category argument (see George Lowther's comment above).
