Yes they can! Every Jordan arc can be included in a Jordan loop. I'll give you the simplest proof I could think of, which only uses very basic topology. It also shows that the lemma you propose does indeed hold, although I'll skip the quantitative aspect. [I see that Jim Conant linked to a very similar question in a comment, but the proof I'll give here is very different from the answers given to the other question. It is also, in my opinion, more elementary, and I prove the lemma asked for here.]

The idea is to use the Jordan arc $\Gamma\subseteq\mathbb{R}^2$ to construct an undirected graph $G_\Gamma$ as follows. Write $S^1=\bar B_1(0)\setminus B_1(0)$ for the unit circle. Then, let the nodes $V_\Gamma$ of the graph consist of the connected components of $\mathbb{R}^2\setminus(\Gamma\cup S^1)$, which naturally split into those inside the circle and those outside. Let the edges $E_\Gamma$ of the graph consist of the connected components of $S^1\setminus\Gamma$. Each edge lies in the closure of precisely two elements of $V_\Gamma$ (one inside and one outside the circle), and we consider it to join these two nodes. Then, the main fact used to prove your lemma is the following.

1) $G_\Gamma$ is a tree.

I'll give a proof of this statement below. As it seems like a rather simple fact, there should be a reference somewhere - maybe some knows one?

Using this, we can easily dispatch with your lemma, which I'll state as follows. I'm using $\Gamma={\rm Im}(\gamma)$ for a continuous one to one map $\gamma\colon[0,1]\to\mathbb{R}^2$.

2) If $\epsilon < 1$ and at least two connected components of $B_1(0)\setminus\Gamma$ meet $B_\epsilon(0)$ then there exists $t_1 < t_2 < t_3\in[0,1]$ with $\Vert\gamma(t_1)\Vert,\Vert\gamma(t_3)\Vert\ge1$ and $\Vert\gamma(t_2)\Vert < \epsilon$.

If the curve $\Gamma$ did not intersect with $S^1$ then $G_\Gamma$ would only have one edge and, hence, only two nodes (the inside and outside of the circle), so only one region could intersect with $B_\epsilon(0)$. On the other hand, suppose that $\Gamma\cap S^1$ is nonempty. Then, let $t_1,t_3\in[0,1]$ be respectively the first and last times at which $\gamma(t)\in S^1$. Let $\gamma^\prime=\gamma\vert\_{[t_1,t_3]}$ and $\Gamma^\prime={\rm Im}(\gamma^\prime)$.
Note that $\Gamma\cap S^1=\Gamma^\prime\cap S^1$, so $E_\Gamma=E_{\Gamma^\prime}$. Also, we can map $V_\Gamma$ to $V_{\Gamma^\prime}$ by taking each connected component $U\in V_\Gamma$ to the connected connected component of $\mathbb{R}^2\setminus(\Gamma^\prime\cup S^1)$ containing it. This is one-to-one on the components outside the circle (as $\Gamma\setminus B_1(0)=\Gamma^\prime\setminus B_1(0)$. By (1), it must be one-to-one on all of $V_\Gamma$. Otherwise, if $U,V\in V_\Gamma$ mapped to the same element of $V_{\Gamma^\prime}$, then a simple path from $U$ to $V$ in $G_\Gamma$ would map to a closed loop in $G_{\Gamma^\prime}$, which must pass through a node outside of the unit circle (as any path alternates between inside and outside the circle), and only passes through this node once (as $V_\Gamma\to V_{\Gamma^\prime}$ is 1-1 on the regions outside the unit circle). This would contradict the statement that $G_{\Gamma^\prime}$ is a tree unless $U=V$. So, $G_{\Gamma^\prime}$ must also have at least two components intersecting with $B_\epsilon(0)$, so $\Gamma^\prime\cap B_\epsilon(0)\not=\emptyset$ and $t_2\in[t_1,t_3]$ exists as required.

As you suggest, this shows that any point $P\in\mathbb{R}^2\setminus\Gamma$ can be connected to $\gamma(0)$ by a by a curve not intersecting with $\Gamma$. Wlog, suppose that $\gamma(0)=0$ and let $t_n\in(0,1]$ be the first time at which $\gamma$ exits from $B_{1/n}(0)$. Then, choose points $P_n\in\mathbb{R}^2\setminus\Gamma$ with $\Vert P_n\Vert\le\min\_{t\ge t_n}\Vert\gamma(t)\Vert$. We can define $f\colon[0,1]\to\mathbb{R}^2$ by letting $f\vert\_{[1/2,1]}$ be a path in $\mathbb{R}^2\setminus\Gamma$ joining $P_2$ to $P$ and, for each $n\ge 2$, let $f\vert\_{[1/{(n+1)},1/n]}$ be a path in $B_{1/n}(0)\setminus\Gamma$ joining $P_{n+1}$ to $P_n$ (which exists, by the lemma). Then, setting $f(0)=0$, this gives the desired curve.

I'll now give a proof of (1), that $G_\Gamma$ is a tree, based on the following facts.

- $\mathbb{R}^2\setminus\Gamma$ is connected.
- If $P,Q,R,S$ are points occuring clockwise (in that order) on $S^1$ and $C_1,C_2\subseteq \bar B_1(0)$ are curves joining $P$ to $R$ and $Q$ to $S$ respectively, then $C_1$ and $C_2$ intersect.

Neither of these facts requires anything particularly difficult. Both of them follow from the methods in the notes *Brouwer's Fixed Point Theorem and The Jordan Curve Theorem* from this link (Section 5) -- see the proof of Lemma 5.6 for the first fact and, reducing to the case $P=(0,1)$, $Q=(1,0)$, $R=(0,-1)$, $S=(-1,0)$, see Lemma 5.5 for the second fact.

The first fact immediately implies that $G_\Gamma$ is connected, so we just need to show that it has no simple closed loops. Suppose, on the contrary, that there was a simple loop with edges $E_1,E_2,\ldots,E_n$ (which are just open segments of $S^1$). Then, as any path alternates between inside and outside the unit circle, $n$ is even. So, the connected components of $S^1\setminus(E_1\cup\cdots\cup E_n)$ consists of $n$ closed connected segments $I_1,\ldots,I_{n/2},J_1,\ldots,J_{n/2}$. I'm labeling these alternately by $I_k$ and $J_k$ as you go round the circle. As each of these closed segments intersects with $\gamma$, there must exist $t_1,t_2\in[0,1]$ with $\gamma(t_1)\in\bigcup_k I_k$ and $\gamma(t_2)\in\bigcup_k J_k$. Wlog, suppose that $t_1 < t_2$. Also, choose $t_1$ maximal in the range $[0,t_2]$ and, then, choose $t_2$ minimal in $[t_1,1]$. Then, $\gamma^\prime=\gamma\vert\_{[t_1,t_2]}$ is a curve joining a point in $I_i$ to $J_j$ (say), and intersects $S^1$ only at its end points. Wlog, we can assume that $\Gamma^\prime={\rm Im}(\gamma^\prime)$ lies in $\bar B_1(0)$ (otherwise, circular inversion can be applied to reduce to this case).

Then, $S^1\setminus(I_i\cup J_j)$ consists of two connected open arcs each of which contains an odd number of the intervals $E_k$. However, the edges of $G_\Gamma$ joining a single node $V\in V_\Gamma$ must all lie in the same arc otherwise, by connectedness of $V$, we would contradict the second fact above. So, each pair of edges $E_k,E_{k+1}$ in the path joining to a node inside the circle lies in the same arc, and hence each arc must contain an even number of the edges $E_k$. This gives the desired contradiction.