Prove every convex set in $\mathbb{R}^k$ is connected.

Question

I have looked up solutions to this when it was asked in Rudin, but the solutions don't really make sense to me, because they use a previous question that Rudin asked as part of the solution. I just want to prove this theorem on its own through connectedness.

The outline of my proof is: if we assume that set E is separated by two sets $A$ and $B$, we let $x\in A$ and $y \in B$. So because it's convex, $\lambda x + (1-\lambda) y \in E$ when $x\in E$, $y \in E$, and $0 < \lambda < 1$. So the set (0,1) is connected in $\mathbb{R}^1$. I'm not really sure where to go from here.

Answer 1

Each two points $x,y$ of a convex set are connected via the path $\alpha x+(1-\alpha)y$.

Answer 2

The way I found makes use of the properties of continuous functions and connected sets.

I assume we're aiming for a contradiction in this. So we suppose the set $E$ is convex and yet not connected, and so equals the union of two disjoint, non-empty open sets $A$ and $B$. Since $E$ is convex, for any $x \in A$, $y \in B$, there exists a mapping $f$ such that \begin{align*} f(\lambda) = \lambda x + (1-\lambda) y \end{align*} for all $\lambda \in (0,1)$. This mapping is continuous and one-to-one, hence its image, which I'll call $P$ for "path", is both connected and open since $(0,1)$ is both connected and open.

Since $A$ and $B$ are disjoint, it follows $P \cap A$ and $P \cap B$ are also disjoint. $P \cap A$ and $P \cap B$ are non-empty because $x$ and $y$ are taken from open sets and so therefore have a "buffer zone" of points around them which $P$ will have to intersect. They are also open sets since $P$ is open and so are $A$ and $B$. Hence $P \cap A$ and $P \cap B$ are two disjoint, non-empty open sets whose union is $P$. Therefore $P$ must be disconnected which contradicts our earlier deduction that it is connected since it is a continuous image of a connected set.