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So far I only know that $S$ is simply connected if it is connected and every loop in $S$ can be shrunk continuously to a point.

And in order to prove simply-connectness, I only have this lemma:

Let $X=U\cup V$, with $U,V$ open and simply connected, and $U\cap V$ is path connected, then $X$ is simply connected.

Then can anyone give me a clue how to prove $\mathbb{R}^3\setminus\{ 0\}$ is simply connected?

Thanks in advance!

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    Can you prove that $\mathbb R^3 \backslash L$ (where $L$ is an half line) is simply connected ? Do you see why this answer your question ? – 2017-02-07
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    I am not very clear how to do that...could you plz give me some move hint? – 2017-02-07
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    Ok first assume what I said is true. Can you use your lemma for finish the proof ? – 2017-02-07
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    Ok,so $\mathbb{R}\setminus \{ 0 \}$ is the union of $\mathbb{R}\setminus L$ and $L\setminus \{ 0\}$. (where $L\setminus \{ 0\}$ is clearly simply connected), so if $\mathbb{R}\setminus L$ is simply connected, then the intersection of those 2 sets are path connected, and therefore we have the desired result? Am I making any sense....thanks! – 2017-02-07
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    Good try but $L$ is not open. Another hint : if you take $U = R^3 \backslash L$, what could be a symmetric choice for $V$ ? – 2017-02-07
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    Sorry...I am getting a little bit lost.. If I take $U=R^3$\ $L$, then in order to fulfill the condition $X=U\cup V$, $V$ must be $L$\{0}? what am I missing here.... – 2017-02-07
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    Ok, take $U = R^3 \backslash L, V = R^3 \backslash (-L)$. Now, you can make a picture and verify all the condition of your lemma :) – 2017-02-07
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    OK! thanks! Now I can see that the union is indeed $R^3$\{0}. But then the intersection of $U$ and $V$ now become the $R^3$\{a whole line in 3-dimensional space}. Then how do I know that the intersection is path-connected? and also how can I show $R^3$\L is open and simply connected? thanks! – 2017-02-07
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    This is correct for the intersection. The remaining steps are easy work that I leave to you as proving my claim in the first line :) If you have difficulties, you can come later here and ask me question again. If this is not immediately clear for you, as I said make a picture and think a bit about it. – 2017-02-07
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    Thanks! The intersection is path-connected is easy indeed. Because after deleting a line in $R^3$, I can still find a continuous path to link any 2 points. (I think it is easy to make sure this path will bypass the line we deleted.) For the first comment you made. I think I should use the lemma again? I will try to prove your comment first and get back to you if I come up with something or some problems! :-) Thank you very much! – 2017-02-07
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    Good for the intersection ! For the first comment I made, I was thinking to retract $R^3 \backslash L$ on a plane for example. For simplicity, you can assume that the $L$ is given by $x=0,y=0, z \geq 1$. Try to retract $R^3 \backslash L$ in the plane $z = 0$ for example :) – 2017-02-07
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    Thanks! Now i meet the new topological term "retract"...after a little bit searching, I am thinking that "retract" somehow means projection? So if we project $R^3$\ $L$ to the xy-plane, we can get $R^2$, which is open and simply connected? – 2017-02-07
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    Ok you are almost done ! You can do without retract but I told you about retract because this is really standard and useful. A retract is an continuous application $r : X \to A$ with $r_{|A} = id$. But this is not very strong : we usually ask a _deformation retract_ (which is a bit like an homotopy). So this exactly your idea, but instead of "projecting" we will slowly bring $R^3 \backslash L$ to the xy-plane. How to do this ? We simply take the application $f_t(x,y,z) = (x,y,(1-t)z)$. This is continuous, maps $R^3 \backslash L$ into itself, and we have $f_1$ = projection, and $f_0 =id$. – 2017-02-07
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    In particular, by this application you have an homotopy between any loop in $R^3 \L$ and the projection on this loop on the $xy$ plane. But the $xy$ plane is simply connected, so your loop becomes zero and you are done ! In general, if you have a deformation retract from $X$ to $A$nyou can deduce that $X,A$ have the same homotopy type which is really useful. For example, $\mathbb C^*$ deform retracts on the circle, etc ... This is not true for a "simple" retraction since you can always take the constant map. You really need this "homotopy" notion. (Sorry to being long !) – 2017-02-07
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    You can think of the following exercise for check if you understood correctly : if $L' \subset R^3$ is a line, is $R^3 \backslash L'$ simply connected ? – 2017-02-07
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    Thank you so much for your detailed reply! Based on your explanations, I do NOT think that $R^3$\ $L$ is simply connected! Also if you are willing to briefly restate your answer (even the hints you gave me) in the Post your Answer. I will choose your answer to be the best! thanks again! :) – 2017-02-08
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    Sure ! And you are right : $R^3 \backslash L$ (where $L$ is a line) is not simply connected, as it deform retract on $R^2 \backslash (0,0)$ which is well-known to be not-simply connected :) – 2017-02-08

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Let $L$ be a half-line. Claim : $U = R^n \backslash L$ and $V = R^n \backslash (-L)$ are simply connected, and $U \cap V = R^n \backslash (L \cup - L)$ is connected, and since $U \cup V = R^n \backslash 0$ we obtain the result.

(In the comments, I took $n=3$ for simplicity but the result holds for any $n \geq 3$.)

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    Here is another proof if you want to get a feeling about it : $R^n \backslash 0$ deform retracts on $S^{n-1}$. So any loop can be bring in $S^{n-1}$, and moreover since $n - 1 \geq 2$ we can assume that it avoid a point (say the North pole). But then we bring our loop in $R^{n-1}$ (this is well known that the point-point compactification of $R^n$ is $S^n$ and $R^n$ is simply connected for any $n$, thus we get again the result. This also show that $S^n$ is simply connected if $n \geq 2$. – 2017-02-08
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    @Wojowu : I don't. The OP stated explicitely a lemma in the beginning (which is of course an immediate corollary of Seifert-van Kampen) and I did use only this lemma. – 2017-02-08
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    Ah, silly me, I have misunderstood what OP meant. From the first sentence I have infered that they don't want to use this lemma, but I see I was mistaken. – 2017-02-08