How to prove $ \mathit {X}$ is path connected?

Question

${\mathbb{R}}^{2} $ Euclidean 2-space,Let $\mathit {X} \subset \mathbb{R}^{2} $.$$\mathit {X}=[-2,2]\times[-1,0]\cup[-2,-1]\times[0,1]\cup[1,2]\times[0,1]$$ $\qquad\qquad\qquad\qquad\qquad$

From the geometric intuition, we know $ \mathit {X}$ is path connected. I want to prove it theoretically.So I tied to use Definition: A topological space $\mathit {X} $ is path connected if for every $x,y\in \mathit {X}$ there is a path in $\mathit {X}$ from $x$ to $y$.but for every $x,y\in \mathit{X}$, finding a continuous function $f:[0,1]\rightarrow \mathit{X}$ such that $f(0)=x$ and $f(1)=y$ is seemingly difficulty. I need some help to go further or some other solutions to prove it. Any of your help will be appreciated!

Answer 1

Hint You can partition your set into three overlapping convex (and hence path connected) subsets, namely three rectangles. Then use that if $A_1,\ldots,A_n$ are path connected and $A_i\cap A_{i+1}$ is nonempty path-connected for $i=1,\ldots,n-1$ then their union is path connected.

Answer 2

Hint:

For any $A(x_1,y_1), B(x_2,y_2) \in X$ there is a path $ACDB$, where $C(x_1;-\frac{1}{2})$ and $D(x_2,-\frac{1}{2})$.

You can try to construct a function $f$ consisting of three parts $f_1:[0,\frac{1}{3}] \to AC$, $f_2:[\frac{1}{3},\frac{2}{3}] \to CD$ and $f_3:[\frac{2}{3},1] \to DB$ (here $f_1(\frac{1}{3})$ should be equal to $f_2(\frac{1}{3})$ and similarly for $f_2$ and $f_3$ in $\frac{2}{3}$).