Show that $E_r(a)=\{x\in\mathbb{R}^n:\|x-a\|>r\}$ is path-connected For $a\in\mathbb{R}^n, r>0,n\ge2,$ show that $E_r(a)=\{x\in\mathbb{R}^n:\|x-a\|>r\}$ is path-connected and hence connected.

So I'm trying to use the idea that for x, y $\in E_r(a)$  if $\exists \varphi(x)\in\mathcal{C}([a,b],E_r(a))$ such that $\varphi(a)=$ x and $\varphi(b)=$ y then $E_r(a)$ is path connected and hence connected.
I figured if I construct a midpoint $z$ using $x,y $ and $a$, and create a piece-wise linear function $\varphi(x)$ between $x$ and $z$ and then $z$ and $y$ I'd be good but the trouble I'm having is constructing the $z$.
 A: Here's a sketch for one possible proof (see also copper.hat's comment).
1) Consider the following ball boundary: $C_{2r}(a) = \{x\in\mathbb R^n \ : \ \|x-a\|=2r\}$.  Show this is path-connected for all $r,a$.  Note: The use of the $2$ here is not important: I just want this set to be entirely contained in the region in question.
2) Given any $x\in E_r(a)$, there exists a line segment $l_x$ with one endpoint equal to $x$ and the other endpoint in $C_{2r}(a)$ and such that $l_x$ is entirely contained in $E_r(a)$.  Denote the endpoint in $C_{2r}(a)$ by $x'$.  Note that $l_x$, and therefore $x'$, is not necessarily uniquely defined, but we do not need uniqueness, only existence.  Drawing a picture may make this fairly clear, but proving it may be slightly trickier.  Try considering segments of minimum length, and observe they must be entirely contained in $E_r(a)$.
3) Given $x,y\in E_r(a)$, pick any segments $l_x$, $l_y$ as above and any path connecting $x'$ and $y'$ in $C_{2r}(a)$.
4) Conclude that the space is path connected.
A: To build on copper.hat's comment and zibadawa timmy's answer, let me fill in a few details.
First of all, let us show that any sphere is path connected.
Let us assume that sphere is centered at the origin, and let $x,y\in S^{n-1}(0,r)$. We surely know how to connect those points with a line, it's just
$$(1-t)x + ty,\quad t\in[0,1]$$ but the problem is that the line obviously isn't contained in the sphere. We can remedy this by normalizing the path and multiply it by $r$ to get path
$$ \alpha(t)=r\frac {(1-t)x + ty}{\|(1-t)x + ty\|},\quad t\in[0,1]$$
Quick check gives us $\alpha(0) = x$, $\alpha(1) = y$ and $\|\alpha(t)\| = r$, unless $x$ and $y$ are antipodal in which case $(1-t)x + ty = 0$ for $t=\frac 12$ and the $\alpha$ is not well-defined. This can be bypassed by choosing any other point on the sphere, and connecting $x$ and $y$ through that point.
Now, for the general case of $x,y\in S^{n-1}(a,r)$ we just have to translate everything to origin, do the same as before and translate it back to get
$$ \alpha_{x,y}(t)=r\frac {(1-t)(x-a) + t(y-a)}{\|(1-t)(x-a) + t(y-a)\|} + a,\quad t\in[0,1]$$ (again, bypassing the problem with antipodal points by choosing another one and connecting through that point.)
Now, choose any sphere that is contained in $E_r(a)$, for example $S^{n-1}(a,2r)$ like in zibadawa timmy's answer. Let us denote it with $S$. Now for any $x\in E_r(a)$, there is a half-ray $r_x$ originating at $a$ and passing through $x$ given by
$$ r_x(t) = t(x-a) + a, \quad t\in[0,+\infty\rangle$$ $r_x(t)$ is in $E_r(a)$ for $t>\frac r{\|x-a\|}$ and intersects $S$ exactly at one point: $t=\frac {2r}{\|x-a\|}$. Denote this intersection with $x'$. Line between $x$ and $x'$ obviously lies in $E_r(a)$ (denote it with $l_x$).
Now remains the obvious: for any two points $x,y\in E_r(a)$, glue together paths $r_x$, $\alpha_{x',y'}$ and $l_y$.
A: The map $\tau_a(x) = x+a$ is a homeomorphism and so maps path connected
sets into path connected sets. Since $E_r(a) = \tau_a ( E_r(0))$, we need only establish that $E_r(0)$ is path connected.
First suppose $x,y \in E_r(0)$ and that $x,y,0$ are not collinear. Let $d$ be the minimum distance from the segment $[x,y]$ to $0$ and note that $d>0$. Let
$t\ge 1$ be such that $t d > r$, then we see that the segment $[tx,ty]$ remains entirely in $E_r(0)$. Then the polygonal path $x \to tx \to ty \to y$
connects $x,y$ in $E_r(0)$.
Now suppose $x,y,0$ are collinear. Let $z \in E_r(0)$ a point that does not
lie on the line through $x,y,0$ (such a point must exist since $n \ge 2$).
Then $x,z,0$ are not collinear and, similarly, $y,z,0$ are not collinear. From
above, we see that there is a path from $x$ to $z$ and a path from $z$ to $y$,
hence there is a path that connects $x,y$ in $E_r(0)$.
A: We need a general principle that produces a path from $x$ to $y$ which avoids the closed ball $B_r(a)$. We may assume $a=0$ and $r<|x|\leq |y|$. 
We now carry out a two-dimensional construction:  Put $e_1:={x\over|x|}$ and choose $e_2\perp e_1$ such that $y\in\langle e_1,e_2\rangle$ and $y\cdot e_2\geq0$. Then there is an $\alpha\in[0,\pi]$ with $y=|y|(\cos\alpha \>e_1+\sin\alpha \>e_2)$. The desired path is then the concatenation of the circular arc
$$\gamma_1:\quad t\mapsto|x|(\cos t \>e_1+\sin t \>e_2)\qquad(0\leq t\leq\alpha)$$ and the segment 
$$\gamma_2:\quad t\mapsto t(\cos \alpha \>e_1+\sin \alpha \>e_2)\qquad \bigl(|x|\leq t\leq |y|\bigr)\ .$$
