Prove that in $\mathbb{R}$ $\text{int([0,1])}=(0,1)$

Question

Let (X,\tau) be a topological space and $A$ any subset of $X$. The largest open set contained in $A$ is called the interior of $A$ and is denoted $\text{Int}(A)$.

i) Prove that in $\mathbb{R}$ $\text{int([0,1])}=(0,1)$

My Attempt:

If $x\in X$ so that $x\in[0,1]$ then I can define $\mathscr{U_x}=(x-\epsilon,x+\epsilon)$ such that so that $\bigcup_{x\in X}\mathscr{U}_x=\text{Int}(A)$ so $\text{Int}(A)$ must be the neighbourhood of all points except of $0$ and $1$ once there is no open set that contains them and is contained in $[0,1]$.

Question:

Is my proof right? If not what would be an alternative one?

Thanks in advance!

Answer 1

(0,1) is an open subset of [0,1],
Subsets larger than (0,1) are [0,1), (0,1], [0,1].
None of those are open.

Answer 2

Note that $(\frac1n,1)\subset [0,1]$ for all $n\in \mathbb N$ and each $(\frac1n,1) $ is open. Since the interior of a set is the largest open subset contained in it then $\cup (\frac1n,1)=(0,1)\subset int[0,1]$. To show that $0,1\not \in int[0,1]$ it is enough to show that there exists no open set $U\subset[0,1]$ s.t $0\in U$ which is clear since $0,1$ are boundary points.