The answers by Kent and Eremenko are, of course, absolutely correct. Here is the more general statement: Let $M$ be a closed connected topological $n-1$-dimensional manifold embedded in ${\mathbb R}^n$ (note that $M$ can be wild for $n\ge 3$). Let $U$ be a component of ${\mathbb R}^n -M$. Then for every point $x\in M$ there exists a continuous path $p: [0,1)\to U$, so that

$$
\lim_{t\to 1} p(t)=x$$

In particular, $({\mathbb R}^n -M) \cup \{x\}$ is path-connected. The proof is based on the following observation:

Observation. There exists a function $\phi(r), r\in [0, \infty)$, satisfying $\phi(r)\ge r$ and

$$
\lim_{r\to 0+} \phi(r)=0$$
so that: For every $r>0$ the map
$$
\tilde{H}_0(U \cap B(x, r))\to \tilde{H}_0(U \cap B(x, \phi(r)))
$$
(induced by inclusion) is zero. In other words, any two points in $U \cap B(x, r)$ can be connected by a path in $U \cap B(x, \phi(r))$.

The above observation follows from the Poincare/Alexander duality in ${\mathbb R}^n$
(in conjunction with the fact that $M$ is a manifold).

Given the above observation, take a sequence $x_k\in U\cap B(x, \frac{1}{k})$ and
construct the path $p$ by concatenating paths $p_i$ in
$$
B(x, \phi(\frac{1}{i}))\cap U
$$
connecting points $x_i, x_{i+1}$, $i\in {\mathbb N}$.