Let $B$ be countable subset of $\mathbb R^n$ where $n > 1$. Prove $\mathbb R^n\setminus B$ is connected space. But I have to show this without arc-connectedness. Hint is to use the fact that sum of connected spaces which have common point is again connected.
To be honest, I don't know how to use the hint. Any help would be great!
First prove that $[0,1]^n\setminus\{p\}$ is connected, where $p$ is arbitrary. This is not difficult, let me know if you need help. Now you can cover $\mathbb{R}^n$ by small enough cubes and use the hint.
to see the connectedness: use induction. I'll leave the proof for $n=2$ to you. wlog assume that $p=(1/2,\dots,1/2)$. Then write $X=[0,1]^n\setminus\{p\}$, and let $$X_1=[0,1]^{n-1}\times[0,1/2]\setminus \{p\}, X_2=[0,1]^{n-1}\times [1/2,1]\setminus\{p\}$$ Then $X$ is a sum of $X_1$ and $X_2$, and $X_1\cong X_2$, so we need only prove that $X_1$ is connected. But we can consider e.g. $$L=\{0\}^{n-1}\times[0,1/2]\cong [0,1/2]$$ Then consider $$X_t=[0,1]^{n-1}\times \{t\}$$ if $t<1/2$ and $$X_{1/2}=([0,1]\setminus\{1/2\})^{n-1}\times\{1/2\}\cong ([0,1]\setminus\{1/2\})^{n-1}$$ using induction this is connected. $X_1=\cup_{t\in[0,1/2]}X_t$ and each of the $X_t$ is connected, and intersects $L$, which is also connected. A standard theorem shows that $X_1$ is connected.