In infinite dimensional vector spaces the complementary of a compact is connected Let $X$ be an infinite dimensional normed vector space and $K$ a compact subset of $X$. I want to show that $K^c$ is connected by path. I know that the unity sphere $\mathbb{S} := S(0,1)$ is connected by path and that up to an homeomorphism I can assume $K \subset \mathbb{B} := B(0,1)$. So I only need to find a path from $x$ to a certain $u \in \mathbb{S}$, for any $x \in K^c$. I was suggested to look at the easiest paths, $t \mapsto (1-t)x+tu$, but I fail to conclude. Any help would be appreciated.

If none of the paths $t\mapsto (1-t)x+tu$ are contained in $K^c$, then for each $u\in\mathbb{S}$ there exists a point of the form $(1-t)x+tu$ in $K$.  In other words, if you could define a continuous map $f:\mathbb{B}\setminus\{x\}\to\mathbb{S}$ that takes $(1-t)x+tu$ to $u$ (geometrically, "project away from $x$ until you hit $\mathbb{S}$") then $f$ would be surjective restricted to $K$ and so $\mathbb{S}$ would be compact, giving a contradiction.
Probably you can do a bit of work to prove this map $f$ is well-defined and continuous.  You can avoid any of that work by ending the argument a different way, though.  In the special case that $x=0$, then $f$ is just the map $v\mapsto v/\|v\|$ which is obviously well-defined and continuous.  So, if you take an arbitrary point $x\in K^c$ and translate it to $0$, this shows that there exists a path in $K^c$ from $x$ to some point in any given sphere centered around $x$.  In particular, taking that sphere to have sufficiently large radius, this gives a path from $x$ to a point that is outside of the ball $\mathbb{B}$.  Since $X\setminus\mathbb{B}$ is path-connected, this completes the proof.
