I want to show that any connected $T_1$ space having more than one point has dimension at least $1$.
Here's my proof:
$X$ is a $T_1$ space i.e if $x,y\in X$ then there exist open nbhds $V_x , V_y$ s.t $x\notin V_y$ and $y\notin V_x$.
Let $\mathcal A := \{ \mathcal{U}_\alpha : \bigcup_\alpha\mathcal{U}_\alpha = X \}$ be an open covering of $X$, then a refinement of it would be: $\mathcal B := \{ \mathcal U_\alpha \cap V_x , \mathcal U_\alpha \cap V_y, V_x ,V_y \}$.
Then since $X$ is connected there exists $\alpha$ s.t $x\in \mathcal U_\alpha \cap V_x$, so we have that every point in $X$ has at least two elements of $\mathcal{B}$ in which it lies in.
Is this right, or do I need to change something in my argument?
Thanks.
HINT: Let $x$ and $y$ be distinct points of $X$, $U=X\setminus\{x\}$, and $V=X\setminus\{y\}$; since $X$ is $T_1$, $\{U,V\}$ is an open cover of $X$. Suppose that $\mathscr{R}$ is an open refinement of $\{U,V\}$.