show that Peano space X is uniformly locally arcwise connected Definition.
A Peano space is a compact, connected, locally connected metric space. 

A Peano space X is uniformly locally arcwise connected;

i.e., 
for each $\epsilon > 0$, there is a $\delta > 0$ such that whenever $\rho(x, y) < \delta$, then $x$ and $y$
are joined by an arc of diameter $<\epsilon$. 
$X$ is Peano space, so $X$ is compact locally connected metric space,
hence, $X$ is uniformly locally connected.
Thus if $\epsilon > 0$ is given, 
there is a $\delta > 0$ such that if $\rho(x, y) < \delta$, then $x$ and $y$ lie together in a connected set $B$ of diameter $<\epsilon/2$.
Since $X$ is locally connected, each $x \in B$ has an open connected nhood $U_x$ of diameter $<\epsilon/4$.
Then $U=\bigcup_{x\in B }{U_x}$ is an open connected subset of $X$.
what could I do after this ?
 A: seem that you read from general topology by Willard, from it and some explanation.
$X$ is Peano space, so $X$ is compact locally connected metric space,
hence, $X$ is uniformly locally connected.
Thus if $\epsilon > 0$ is given, 
there is a $\delta > 0$ such that if $\rho(x, y) < \delta$, then $x$ and $y$ lie together in a connected set $B$ of diameter $<\epsilon/2$.
Since $X$ is locally connected, each $x \in B$ has an open connected nhood $U_x$ of diameter $<\epsilon/4$.
Then $U=\bigcup_{x\in B }{U_x}$ is an open connected subset of $X$.
(To see this, since $B$, $U_x$ are connected sets such that $B \cap U_x \ne \phi$, because $x \in B\cap U_x$. Thus $B \cup U_x$ is conected by using Theorem 26.7 part (c).\ Now $U=\bigcup_{x\in B }{U_x}=\bigcup_{x\in B }({U_x \cup B})$, where $U_x \cup B$ are connected and $\cap( U_x \cup B )\ne \phi $ because $B \subset \cap( U_x \cup B )$.
Thus $U=\bigcup_{x\in B }{U_x}$ is connected by using Theorem 26.7 part (a) .)
Hence, (see Exercise 31C.1), $U$ is arcwise connected.
Thus, if $\rho(x, y) < \delta$, then $x$ and $y$ lie in an arcwise connected subset $U$ of diameter $<\epsilon$.
(To see this, let $a$, $b$ $\in U=\bigcup_{x\in B }{U_x}$, and $a \in U_{x1} $, $b \in U_{x2}$ where $x_1, x_2 \in B$.
So, $\rho(a, b) < \rho(a, x_1)+\rho(x_1, x_2)+\rho(x_2, b)<\epsilon/4 + \epsilon/2 + \epsilon/4 =\epsilon $)
