Prove that $\mathbb R^2$ is connected I tried to prove that $\mathbb{R}^2$ is connected with the usual euclidean metric. There is a hint: Consider subspaces of the form $\{a\}\times \mathbb{R}$ and $\mathbb{R}\times \{b\}$. This makes confuse. Anyone can help me with this question? Prove $\mathbb{R}^2$ is connected. Thanks.
 A: Recall that $\mathbb R$ is connected, and notice that it is homeomorphic to vertical and horizontal slices of the form $\{a\} \times \mathbb R$ and $\mathbb R \times \{b\}$ so that these slices are also connected. Now fix the base point $(0, 0) \in \mathbb R^2$ and for any $a \in \mathbb R$, consider the family of cross-shaped spaces of the form:
$$
T_a = (\{a\} \times \mathbb R) \cup (\mathbb R \times \{0\})
$$
By taking the union of all such cross-shaped spaces over all $a \in \mathbb R$, we obtain the entire plane. Hence, since $\mathbb R^2$ is the union of a collection of connected spaces that have a base point $(0, 0)$ in common, we conclude that $\mathbb R^2$ must also be connected, as desired. $~\blacksquare$
A: If $(x,y),(u,v)\in\mathbb{R}^2$ then the function $f:[0,1]\to \mathbb{R}^2$ defined by $$f(\xi ) =\xi (u,v) +(1-\xi )(x,y)$$ is continuous and $f(0) =(x,y) , f(1) =(u,v).$ Hence the space $\mathbb{R}^2$ is path connected, but every path connected space is connected.
A: Hint: Prove that $\Bbb{R^2}$ is path connected. That is there is always a path $\alpha$ (continuous map) joining two points in $\Bbb{R^2}$. Now since we know that every path connected set is connected your $\Bbb{R^2}$ is connected
A: For an approach that uses the hint you were given, fix $a,b\in \mathbb R$. 
Now take $A_x=\mathbb R\times \left \{ b \right \}\cup \mathbb R\times \left \{ x \right \}$. This set is connected, being the union of two connected sets (lines, which are homeomorphic to $\mathbb R$) and have a point in common.
To finish, note that $\mathbb R^2=\bigcup _{x\in \mathbb R}A_{x}$ is connected because it is a union of sets all of which have a point-namely, $(a,b)$- in common. 
A: Recall that $[0,1]$ is connected. A construction of Peano gives a space-filling (read: surjective) continuous curve $f: [0,1] \to [0,1] \times [0,1] \subset \mathbb{R}^2$. Since continuous images of connected sets are connected, we deduce $[0,1] \times [0,1]$ is connected.
There is clearly a homeomorphism between $[0,1] \times [0,1]$ and any closed rectangle in $\mathbb{R}^2$. Hence, any closed rectangle in $\mathbb{R}^2$ is connected.
Let $R_n = [-n,n] \times [-n,n]$. Since $R_n \subset R_{n+1}$, any two of the rectangles $R_i$ have nonempty intersection, thus the union of all $R_i$ is connected. But $\bigcup_{i=1}^{\infty} R_i = \mathbb{R}^2$. We conclude $\mathbb{R}^2$ is connected.
Since space-filling curves exist in all dimensions, this proof can be extended to show $\mathbb{R}^n$ is connected for all $n\ge 2$.
