Show that from a system of closed intervals covering an open interval it is not always possible to choose a finite subsystem covering the interval

Show that

a) from a system of closed intervals covering a closed interval it is not always possible to choose a finite subsystem covering the interval;

b) from a system of open intervals covering an open interval it is not always possible to choose a finite subsystem covering the interval;

c) from a system of closed intervals covering an open interval it is not always possible to choose a finite subsystem covering the interval.

---- Zorich Mathematical Analysis I 2.3 Exercise 2

My attempt for all of the questions is the following:

Proof for a)

Let$\bigcup_{n=1}^{\infty}I_n=\bigcup_{n\in\mathbb N}{[0,1-\frac 1 n]}=[0,1]$For any $x\in(0,1]\subset\mathbb R,\:\exists n\in\mathbb N\:$such that$\: nx>1$But$\:0\:\notin (0,1],$And that $\:0\cdot n=0<1\:\forall n\in N$

It is shown that there isn't a finite subsystem covering the interval.

Therefore from a system of closed intervals covering a closed interval it is not always possible to choose a finite subsystem covering the interval.

Proof for b)

Let $\bigcup_{n=1}^{\infty}I_n=(\frac 1 2,1)\cup\bigcup_{n\in\mathbb Z_{\ge 2}}(0,\frac 1 2-\frac 1 {n+1})=(0,1)$

For any $$x\in (0,1)\implies x\in(\frac 1 2,1)\vee x\in(0,\frac 1 2-\frac 1 3)\vee x\in(0,\frac 1 2-\frac 1 4)\vee\cdots$$$x=\frac{1} {2}\in (0,1)\quad$But $x$ cannot be covered by finite subsystem

Therefore a system of open intervals covering an open interval doesn't always have a finite subsystem covering the interval.

Proof of c)

Let$(a,b)=\bigcup_{n=1}^{\infty}{[a+\frac{1} {n},b-\frac{1} {n}]}$Because $a,b\notin (a,b)$, there aren't a finite subsystem covering the open interval.

I think my proof and construction is unrigorous, and especially for c).

The reason why is for I think that my proof for c) is that how can we prove that there aren't finite subcoverings covering the range $(a,b)$ when as $I_n$'s n progresses, it is covering more and more of $(a,b)$. Can't I just choose the greatest n and call that a finite covering and call it a day?

So may someone help me to give a more rigorous proof and also I noticed that these concepts are from topology, so maybe can you tell me what concepts are these corresponding to, and in order to give a formal proof. Thank you.

Edit:

1. It seems that the proof of a) is wrong, since that $\bigcup_{n=1}^{\infty}I_n=\bigcup_{n\in\mathbb N}{[0,1-\frac 1 n]}=[0,1).$
2. It seems also that the proof of b) is wrong,since that $\bigcup_{n\in\mathbb Z_{\ge 2}}(0,\frac 1 2-\frac 1 {n+1})\ne (0,\frac 1 2]$

Edit 2:Seems like for b) We can let $\bigcup_{n=1}^{\infty}I_n=(0,a_1]\cup (a_1,a_2]\cup\cdots$for all $a_i<a_{i+1},a_i<1$

And for a) we can use $\bigcup_{x\in(0,1)}[x,x]$