Prove $[1,2)$ homeomorphic to $(-1,0]$ Prove $[1,2)$ homeomorphic to $(-1,0]$

Proof: Consider the function $$f(x)=1-x$$

This function is continuous. Because if an $S \subset (-1,0]$ open in $(-1,0]$
Then $$f^{-1}(S)=(x \in [1,2)  |f(x) \in S)$$ is open in $[1,2)$ since for every $x\in f^{-1}(S)$ i define the hood of $x \in U_x=(x-δ_x,x+δ_x)$ where $δ_x=1-δ_s$ where $δ_s$ is the number that for every $s \in S$ i know $(s-δ_s,s+δ_s) \subset S$ Now im stuck at proving that $U_x \subset f^{-1}$ for every $x$.I know it is open.I can prove the rest . 1-1 ,onto.ALso i might stuck on proving that the inverse function is continuous since if i try the same way 
 i ll stuck at the same step!! 

So if a $x \in U_x$ have to prove that $f(x) \in S$. I know $x-δ_x&lt;x&lt;x+δ_x$.THats as far as i go.Maybe i can use calculus and say that since my function is increasing  then $f(x-δ_x)&lt;f(x)&lt;f(x+δ_x)$ But i dont know if i can do that since my topological spaces are not the usual $R$ .And using caulculus seems maybe odd? My first time taking topology in a semester so not sure what can i use and what i can do. So i need guidenance  to complete the proof.

I dont want some other way.You can add other ways but they wont be considered as answers.Except if my way is wrong!!
Let $f(x):([1,2),\mathcal{T_{[1,2)}}) \to ((-1,0],\mathcal{T_{(-1,0]}})$ such that $f(x)=1-x$ which is the function mentioned in the comments.

The above topologies are the topologies induced  on $[1,2)$ and $(-1,0]$ respectively from the usual topology of $\Bbb{R}$

This function is a bijection thus $f[1,2)=(-1,0]$


  A function $f:X \to Y$ where $X,Y$ are topological spaces is continuous at a point $x$ if fro every open  $V \subseteq Y$ containig $f(x)$,exists and open $U \subseteq X$ such that $x \in U$ and $f(U) \subseteq V$


This is the definition of continuity between two topological spaces.

Now let $x_0 \in[1,2)$ and $V$ an open set in $(-1,0]$ such that $y_0=f(x_0) \in V$

From the properties of a topological subspace we have that $V=(-1,0] \cap O$ where $O$ is open in $\Bbb{R}$ w.r.t the usual topology.

So $y_0 \in O$ and since $O$ is open in $\Bbb{R},$

exists $\epsilon&gt;0$ such that $y_0 \in (y_0-\epsilon,y_0 +\epsilon) \subseteq O$ and also $$\{x \in [1,2)|f(x) \in (y_0-\epsilon,y_0+\epsilon)\}=f^{-1}(y_0 -\epsilon,y_0+\epsilon)=(x_0-\epsilon,x_0+\epsilon) \Rightarrow (y_0 -\epsilon,y_0+\epsilon)=f(x_0-\epsilon,x_0+\epsilon)$$ since $f$ is a bijection.

Now consider the set $U=[1,2) \cap (x_0-\frac{\epsilon}{2},x_0+\frac{\epsilon}{2})$

We have that $U$ is open in $[1,2)$ and contains $x_0$ and 

Also $$f(U)=f[1,2) \cap f(x_0-\frac{\epsilon}{2},x_0+\frac{\epsilon}{2}) \subseteq f[1,2) \cap f(x_0-\epsilon,x_0+\epsilon)$$ $$=(-1,0] \cap (y_0-\epsilon,y_0+\epsilon)$$ $$\subseteq (-1,0] \cap O=V$$

We proved that $f$ is continuous at $x_0$ w.r.t the induced topologis on these intervals and since $x_0$ was arbitrary we have that $f$ is continuous on $([1,2),\mathcal{T_{[1,2)}})$

The inverse of $f$ is $g(y)=1-y$.

With the same exact argument you can prove that $g$ is continuous,thus $f$ is a homeomorphism between the above spaces.
Consider $f(x)=-x+1$.
Where $f :[1,2)\to (-1,0]$.