> Let $E\subsetneq \mathbb{R}^n$ be a vector space. $\mathbb{R}^n-E$ is connected $\iff dim(E) \leq n-2$ 
Let $E\subsetneq \mathbb{R}^n$ be a vector space. $\mathbb{R}^n-E$ is connected $\iff dim(E) \leq n-2$

I was able to prove one way, I couldn't make the other.
$(\Rightarrow)$ Let $\mathbb{R}^n-E$ be connected. Suppose $dim(E)>n-2 \implies dim(E) = n-1$ since $E$ is not $\mathbb{R}^n$.
Let $B =\{b_1,...,b_{n-1}\}$ orthonormal base of E and $B\cup \{b_n\}$ an orthonormal base for $\mathbb{R}^n$. 
$\forall x \in \mathbb{R}^n-E$, $x = \sum_{i=1}^n a_ib_i$, $a_n\not =0$. (if $a_n=0$ then $x \in E$)
Let $A = \{y \in \mathbb{R}^n| y =\sum_{i=1}^n a_ib_i$, $a_n>0\} $
Let $C = \{y \in \mathbb{R}^n| y =\sum_{i=1}^n a_ib_i$, $a_n<0\} $
$\mathbb{R}^n-E = A \cup C$, $A$ and $C$ are open and $A \cap C = \emptyset$ therefore $\mathbb{R}^n-E$ isn't connected. Contradiction! So $dim(E)\leq n-2$.
Now, what I was attempting for the other part:
($\Leftarrow$)Suppose $dim(E)=m\leq n-2$
Let $B = \{b_1,..,b_m\}$ orthonormal base of $E$ and $B^*=B\cup\{b_{m+1},...,b_n\}$ orthonormal base of $\mathbb{R}^n$.
Suppose $\mathbb{R}^n-E = A \cup C$, $A,C$ open sets and $A \cap C = \emptyset$. 
I need either to show that A (or C) is an empty set or suppose there's an element in both sets A and C and get to a contradiction. This part I couldn't find a way. Any useful hints?
ps:The first part is correct?
 A: If $\text{dim}(E) \leq n - 2$, we can find a linearly independent set $\{v_1,v_2,\ldots,v_k,u_1,u_2\}$, where $\{v_1,\ldots,v_k\}$ is a basis for $E$. Let $W_1 = \text{Span}(v_1,v_2,\ldots,v_k,u_1)$, and $W_2 = \text{Span}(v_1,v_2,\ldots,v_k,u_2)$.
Now for any vector $x \in \mathbb{R}^{n} \backslash E$, if $x \notin W_1$, then the straight line connecting $x$ and $u_1$ lies in $\mathbb{R}^{n} \backslash E$, so $x$ and $u_1$ lie in the same connected component of $\mathbb{R}^{n} \backslash E$. Similarly, if $x \notin W_2$, then $x$ and $u_2$ lie in the same connected component of $\mathbb{R}^{n} \backslash E$. And the straight line between $u_1$ and $u_2$ is also in $\mathbb{R}^{n} \backslash E$, so these connected components are the same, and comprise all of $\mathbb{R}^{n} \backslash E$. 
A: Assume $dim(E) \leq n-2$
Here is the solution for Dim($E$) = 1 and n = 3.
Use the standard $x,y,x$ coordinate system with $E$ the $z$-axis.
The set $\{v = (x, y, z) \text{ such that } x^2 + y^2 \gt 0\}$ is the complement of $E$ and is connected since you can path connect any point outside of $E$ to $(1,0,0)$.
A: Suppose that the $\mathrm{Dim}(E) \leq n-2$. Then, write down a basis $e_1,e_2, \dots,e_{n-2}$ for $E$, and extend to a full basis by $e_{n-1}, e_n$. Then, given any points  $x,y \in \mathbb R^n-E$, we have that $x=a_{n-1} e_{n-1}+a_ne_n$ and $y=b_{n-1} e_{n-1}+b_n e_n$. In this case, we can construct the line $y-x \in \mathbb R^n-E$, which supplies a perfectly good path between the points, assuming that it does not pass through the origin. From this, it follows that the subspace is path connected, and hence connected.
Explicitly: $f:[0,1] \to \mathbb R^n-E$ given by $f(t):=x+ty=(a_{n-1}+tb_{n-1})e_{n-1}+(a_n+tb_n)e_n$ gives a path.
If the line paases through the  origin, we need only take $c:=(1,\dots,0)$ and construct a path piece-wise through $c$.
