Let $X = [0,3] \subset\mathbb{R}$ and consider the following equivalence relation: $$ x\sim y \Leftrightarrow x=y \vee x,y \in [1,2]$$ and call $Y = X/\sim$.
(1). Establish if $Y$ is connected, compact, Hausdorff.
(2). Establish if X is homeomorphic to Y.
Proof.
For the point (1), I know that $Y$ is connected, compact because $X$ is compact and connected and
$$\pi:X\rightarrow\mathit(X/\sim)$$
is continuous and surjective.
I think that $Y$ is also Hausdorff because if
$x$, $y\in [0,1)\cup(2,3]$
$X$ is Hausdorff, so I can find $\mathcal{U}_{x}$, $\mathcal{U}_{y}$ such that $\mathcal{U}_{x}\cap\mathcal{U}_{y}=\emptyset$.
Now I can choose
$\mathcal{V}_{x}=\pi(\mathcal{U}_{x}\cap(X\setminus[1,2])$
$\mathcal{V}_{y}=\pi(\mathcal{U}_{y}\cap(X\setminus[1,2])$.
If $x\in[0,1)\cup(2,3]$ and $y\in[1,2]$ I can choose
$\mathcal{U}_{x}$, $\mathcal{U}_{1}$ disjointed
$\mathcal{U'}_{x}$, $\mathcal{U}_{2}$ disjointed too
Now I can choose
$\mathcal{V}_{x}=\pi(\mathcal{U}_{x}\cap\mathcal{U'}_{x})$
$\mathcal{V}_{y}=\pi(\mathcal{U}_{1}\cup\mathcal{U}_{2})$
If $y\in[0,1)\cup(2,3]$ and $x\in[1,2]$ I can do the same
If $x$, $y\in[1,2]$ we have that
$\pi(x)=\pi(y)$
So we have that for each $x$, $y\in Y$ I can find $\mathcal{V}_{x}$, $\mathcal{V}_{y}$ open sets disjointed.
Now I don't know how to continue for the point (2).
Can someone help me?
I'll give you a mental image to help you understand what the homeomorphism should look like:
Since you identify the closed interval $[1,2]$, you can think about this as taking the interval $[0,3]$, and "squishing" its middle part (the interval $[0,2]$) into a single point. This basically leaves you with a "smaller" version of $[0,3]$, which is only of length $2$.
Can you think of some nice $f$ that will display the procedure of "squishing" the interval?
(A little tip - if $f:X\rightarrow Y$ is continuous and bijective, $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism. Meaning - you don't need to make sure that the inverse is continuous )