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If the answer is no: what does it happens if the space in wich lies the point is connected?

PS: sorry for my bad English.

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    By definition of a *neighbourhood*, I think it is already open. – 2017-02-09
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    Well, at least my definition is: a neighborhood of point is a set containing an open subset that contain the point, so no, it isn't already open... – 2017-02-09
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    I see. This definition is different from mine. But I love the answer below. – 2017-02-11
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    @астонвіллаолофмэллбэрг. I have noticed that on this site the convention of "neighborhood" is the def'n in the comment by Davide F. Also, on this site $\mathbb N$ denotes the positive integers, i.e. $0\not \in \mathbb N,$ which I have reluctantly accepted. – 2017-06-18
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    @DanielWainfleet Sorry for the late reply, but I suppose we can't do much about that. Conventions even depend upon where you have done your mathematical education, so occasionally when I do find contrasting notions and definitions, I feel the same way you do. In fact, up till now, whenever I have used $\mathbb N$, I have always specified that it contains zero. Most people on this site don't include it, the set with zero is called "whole numbers", I believe. So I suppose we can reluctantly accept some things. Which is fine for me, since everything else is fun on this site. – 2017-06-27

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Here's a counterexample. Let $X\subset\mathbb{R}^2$ be the topologist's sine curve $\{0\}\times[-1,1]\cup\{(x,\sin(1/x)):0Warsaw circle. Let $Z$ be an arc that oscillates and approaches a segment of $Y$, so the union of $Z$ and that segment is another topologist's sine curve, and consider the space $A=X\cup Y\cup Z$.

Note that $A$ is connected, since it is the union of the two connected sets $X\cup Y$ and $\overline{Z}$ which have nonempty intersection. The point $(0,0)\in A$ has a path-connected neighborhood in $A$, namely the set $X\cup Y$. But $(0,0)$ has no path-connected open neighborhood in $A$: it is easy to see a path-connected neighborhood of $A$ must contain all of $X\cup Y$, and therefore must intersect $Z$ if it is open, but no path in $A$ can have points in both $Z$ and $X\cup Y$.