Show that any convex subset of $R^k$ is connected I need to prove that any convex subset of $R^k$ is connected. I have seen the proof in Rudin's book and on numerous websites but they all use some prior results. I want to do it without using results where one establishes something for some other question and then uses the result to prove this.
 A: First a preliminary lemma:
Lemma: If $\mathcal{F}$ is a family of connected sets such that $\bigcap \mathcal{F} \neq \emptyset$ then $\bigcup \mathcal{F}$ is connected, for, if $\bigcup \mathcal{F}$ is not connected, then it can be written as the union of two disjoint open subsets $X$ and $Y$. Let $x \in \bigcap{F}$. Suppose that $x \in X$, so if $y \in Y$ then there is $F \in \mathcal{F}$ such that $y\in F$ so $F = (A\cap F) \cup (B\cap F)$, wich is a contradiction, since $A$ is open in $\bigcup \mathcal{F}$, $A \cap F$ is open in $F$ and $B \cap F$ is open in $F$.
Now, the problem:
Let $X\subset M$ be a non-empty convex subset and let $a \in X$. Let $L_v = \{x \in X \ |\  x = y + \lambda \ v$ for some $\lambda \in \mathbb{R} \}$. As $X$ is convex, $$X = \bigcup_{\|v\|=1} L_v$$
It remains to prove that each $L_v$ is connected, but since $L_v$ is homeomorphic to some real interval (because $X$ is convex!), $L_v$ is also connected. And since $\{a\} \subset \bigcap_{\|v\|=1} L_v $, lemma aplies and thus $X$ is connected.
