For a family of sets {$A_n $}$_{n\in\mathbb{N}}$ given by $A_n=(0,n)$ prove that $\bigcap_{n\in\mathbb{N}}A_n=(0,1)$ [closed]

The set of subsets of a partricular set $S$ defined as {$A_i: i \in I$} where $\forall i \in I, A_i$ is a unique subset of $S$, is called the family of subsets of $S$ indexed by $I$. In this case $I\neq\phi$ is called the index set of the family of sets.

Intersection of the elements in the family denoted by $\bigcap_{i\in I}A_i$ is given by,

$\bigcap_{i\in I}A_i=${$x\in S: \forall i \in I, x\in A_i$}

There is another property taught by our teacher which states that

$X=\bigcap_{ i\in I}A_i \iff$(($\forall i \in I, X\subseteq A_i$) and ($Y\subseteq A_i,\forall i \in I \implies Y \subseteq X$))

Now comes the solution part.

$\forall n \in \mathbb{N}, n \geq 1$ and $1>0$

$\therefore 0<1<n,\forall$($n\in\mathbb{N}$ and $n>1)$

Now, property of interval states that for an interval $(x,y),x<z<y\implies (x,z]\subseteq(x,y)$ and $[z,y)\subseteq(x,y)$

Hence, $\forall$($n\in\mathbb{N}$ and $n>1), (0,1]\subseteq A_n$

Now, $(0,1) \subseteq (0,1]$

$\therefore\forall(n\in\mathbb{N}$ and $n>1),(0,1)\subseteq A_n$

Again, $(0,1)\subseteq(0,1)$ [Equal sets are subsets of each other]

i.e. for $n=1\in\mathbb{N}$ also $(0,1) \subseteq A_n$

$\therefore\forall n\in\mathbb{N},(0,1)\subseteq A_n$

I'm a bit confused about proving the latter part. Can anyone help me figure it out or suggest me a better approach to the question? Am I missing something while figuring out the solution?