Study some topological properties of $I^{\aleph_0}\times I^2/M$

I've been solving some problems from my Topology course, and I'm unable to finish this one:

Given $I^{\aleph_0}$ the Hilbert cube, and given $Y=I^2/M$ the quotient space where $M=\{(0,0),(1,0),(0,1),(1,1)\}$.

1. Prove that $X=I^{\aleph_0}\times Y$ is $T_1$, compact and path connected.

2. Prove that given $f:X\to\mathbb R$ continuous map, $\exists a,b\in\mathbb{R}$ s.t. $a\leq b$ > and $f(X)=[a,b]$ (assuming that $\mathbb R$ has the usual topology).

The work I've done:

Question 1

• Compact: since $I^{\aleph_0}$ and $I^2$ are topological cubes, they're compact. $Y$ is then also compact because it's the continuous and surjective image of a compact (for being a quotient space). Hence $X$ is compact because it's product of compact spaces.
• Path connected: using an analogous reasoning as the one used for compactness (cubes are path connected, quotient of path connected is path connected, and product of path connected is path connected) it follows that $X$ is path connected.
• $T_1$: here's where I have difficulties in this question. I know cubes are $T_2$ and also that they're $T_4$ (normal + $T_1$). I would be able to prove that $Y$ is normal since the quotient map defined is closed (easy to see because $M$ is finite, hece closed for cubes being $T_1$, and if the subset below is closed then the quotient application is also closed), but still proving that $Y$ is normal does not seem useful to prove that $X$ is $T_1$. I'm probably forgetting some characterisation property that may turn this prove into a trivial one (since using the definition in a space as complicated as this one seems unachievable and clearly not the exercise's intention), but right now I don't know how to prove it.

So here's my first question: How can I prove this space $X$ is $T_1$? Which characterisation am I forgetting?

Question 2

For this second one, I assume I've proven the first.

• From my course notes, I know a result that tells me that if $X$ a non-empty compact space, then a continuous map $f:X\to\mathbb R$ is bounded.

• I also know that the continuous image of a path connected space is also path connected, and a subset of usual $\mathbb R$ is path connected if and only if it's an interval. Considering also that it's bounded, then we conclude $\exists a,b\in\mathbb R$ s.t. $f(X)$ is of one of the following forms: $(a,b)$, $(a,b]$, $[a,b)$ or $[a,b]$.

• The continuous and surjective image of a compact set is compact. If I consider $f:X\to f(X)\subset \mathbb R$, it's clearly continuous and surjective, hence we conclude $f(X)$ is compact, and since it's a subspace of usual $\mathbb R$ then it's closed and bounded. From the 4 different types of interval that $f(X)$ could be as seen in the previous point, the only one closed is $[a,b]$.

So from this reasoning I can finnaly conclude $f(X)=[a,b]$ for some $a,b\in\mathbb R$, $a\leq b$. I feel this prove is correct but maybe I jumped into a wrong conclusion somewhere midway. So my second question is : Is my solution for the second question of the exercise correct? If not, where did I made a mistake?

My biggest issue with this problem is proving $X$ is $T_1$ in question 1, for question 2 I just want to check if my proof is correct. Any help ot hint will be appreciated, thanks in advance.

1. The preimage of any point by the quotient map is a subset of at most four points, and therefore a closed set. Thus, every point in the quotient is closed.

2. $X$ is compact, path connected and non-empty, therefore the image of $f$ must be a non-empty, compact and connected subset of $\Bbb R$, id est an inteval in the form $[a,b]$ for some $a\le b$. Personally, I would just say that connected subsets of $\Bbb R$ are intervals and that compact subsets of $\Bbb R$ have a maximum and a minimum, but your work doesn't seem incorrect.