If $X$ is path connected then $X$ has not isolated point. 
Lemma
If $X$ path connnected then for any $x_0$ there exist a path such that connect $x_0$ with any other point $X$ of $X$.
Statement
If $X$ is path connected then $X$ has not isolated point.

Unfortunately I can't prove the statement, but I'm sure that it is posisble to prove it showing that if $x_0\in X$ is an isolated point for $X$ then there aren't continuous path that connect $x_0$ with any other $x\in X$. If the statement is generally false then is it true for $\Bbb{R}^n$? So could someone help me, please? 
 A: The statement is not quite true: if $X=\{x_0\}$, then $X$ is path connected, and its only point is isolated. It can also fail if $X$ is not $T_1$. Let $X=\{0,1\}$ with open sets $\varnothing$, $\{0\}$, and $\{0,1\}$, and let
$$f:[0,1]\to X:x\mapsto\begin{cases}
0,&\text{if }0\le x<\frac12\\
1,&\text{if }\frac12\le x\le 1\;.
\end{cases}$$
Then $0$ is an isolated point of $X$, $f(0)=0$, $f(1)=1$, and $f$ is continuous, because $f^{-1}[\varnothing]=\varnothing$, $f^{-1}[\{0\}]=\left[0,\frac12\right)$, and $f^{-1}[X]=[0,1]$ are all open in $[0,1]$.
It is true, however, if $X$ is $T_1$ and has at least two points. Suppose that $x_0\in X$ is isolated, and $x_1\in X\setminus\{x_0\}$; we’ll prove that there is no path from $x_0$ to $x_1$. 
Suppose that $f:[0,1]\to X$ is a continuous function such that $f(0)=x_0$ and $f(1)=x_1$. Let $U=f^{-1}[\{x_0\}]$; $\{x_0\}$ is both open and closed in $X$, so $U$ is both open and closed in $[0,1]$. Let $V=[0,1]\setminus U$; $V$ is also both open and closed in $[0,1]$, and since $1\in V$ (why?), $V\ne\varnothing$. But $[0,1]$ is connected, so no such separation can exist. Thus, there is no such function $f$, and $X$ is not path connected.
A: A space with two or more points that is $T_1$  and connected, cannot have an isolated point. Proof: otherwise the singleton of that point is nontrivial clopen. The separation axiom is needed: a two point Sierpi&nacute;ski space is path-connected but has an isolated point. 
