Is the compact complement topology locally arc connected? Let $\tau$ be the standard topology on $\mathbb{R}$ and let $\tau'$ be the compact complement topology on $\mathbb{R}$: $$\tau' = \{\mathbb{R} \setminus K \mid K \text{ is $\tau$-compact}\} \cup \{\varnothing, \mathbb{R}\}$$
π-Base claims that $(\mathbb{R},\tau')$ is locally arc-connected. I believe the proof that it is arc-connected: since $\tau' \subseteq \tau$, the identity map $id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R},\tau')$ is a continuous injection and thus gives an arc between any two distinct points. 
They give the same proof for local arc-connectedness, but it is insufficient since we must show that every nonempty open set contains an arc-connected open neighborhood of each of its points. In particular, I'm skeptical that any nonempty, proper, $\tau'$-open set can even be path-connected. 
 A: You are correct that the given argument does not work.  However, $(\mathbb{R},\tau')$ is in fact locally arc-connected.  It suffices to show that if $a<b<c<d$ then the open set $U=(-\infty,a)\cup(b,c)\cup(d,\infty)$ is arc-connected, since such open sets form a basis.  I will show that if $x\in (d,\infty)$ and $y\in (b,c)$ then there is a continuous injection $f:[0,1]\to U$ with $f(0)=x$ and $f(1)=y$; the other cases are similar.
To define $f$, first let $g:[0,1)\to[x,\infty)$ be an order-isomorphism.  Then define $f(t)=g(t)$ for $t\in[0,1)$ and $f(1)=y$.  Clearly $f$ is injective, and it is continuous at any $t\in[0,1)$ since $g$ is continuous with respect $\tau$ and hence also with respect to $\tau'$.  So we just need to check continuity at $1$.  If $V\subseteq U$ is any $\tau'$-open set containing $y$, then $V$ must contain $(z,\infty)$ for some $z\in[x,\infty)$.  Thus $f^{-1}(V)$ contains the entire interval $(g^{-1}(z),1]$, and in particular contains a neighborhood of $1$.  Thus $f$ is continuous at $1$.
