Every finite connected space is also path-connected? Let  $X$ be a connected space, if $X$ is finite, then $X$ is a path-connected space? If so, how to prove it? If not, how to give a counterexample?
Thanks in advanced.
 A: For $x \in X$, let $U_x := \bigcap \{U \subseteq X : U \text{ open}, x\in U\}$ denote the smallest open set containing $x$. 

Lemma. Let $x, y \in X$, then there is a sequence $x_0 = x, x_1, \ldots, x_n= y$ such that for each $i$, either $x_i \in U_{x_{i+1}}$ or $x_{i+1} \in U_{x_i}$.

Proof. For $x \in X$, let $A \subseteq X$ denote the set of points $y\in X$ for which such a sequence exists. We have $x \in A$, and for any $y \in A$, we have $U_y \subseteq A$, hence $A$ is open. If $z \not\in A$, then $U_z \subseteq X \setminus A$, hence $A$ is closed. As $X$ is connected, $A=X$.

Lemma 2. If $x \in U_y$, then there is a path connecting $x$ and $y$. 

Proof. Define $w \colon [0,1] \to X$ by $w(t) = x$ for $t < 1$ and $w(1) = y$. Let $U \subseteq X$ open, if $y \not\in U$, then $w^{-1}[U] \in \{\emptyset, [0,1)\}$, hence $w^{-1}[U]$ is open, if $y \in U$, then $U_y \subseteq U$, hence $x \in U$, therefore $w^{-1}[U] = [0,1]$. Therefore $w$ is continuous.

Proposition. $X$ is path-connected.

Proof. Let $x,y \in X$, by Lemma 1 there is a sequence $x_0, \ldots, x_n$ with $x_0 = x$, $x_n = y$, and $x_i \in U_{x_{i+1}}$ or $x_{i+1} \in U_{x_i}$ for all $i$. By Lemma 2, $x_i$ and $x_{i+1}$ are connectable by a path $w_i$. 
As this is true for all $i$, $x_0$ and $x_n$ are connected by a path.
