Proving that $\tau = \{N,\varnothing\}\cup \{S_n:n\in\mathbf{N}\}$ is a topology on $\mathbf{N}$. I am required to show that $\tau = \{N,\varnothing\}\cup \{S_n:n\in\mathbf{N}\}$ where $S_n := \{1,\dots,n\},\forall n\in\mathbf{N}$ is a topology on $\mathbf{N}$, the part where i am experiencing some dfficulty is in proving that the union of a countably infinite subset of $\{S_n:n\in\mathbf{N}\}$ belongs to $\tau$.
Here is my attempt so far, Is it correct?
Let $\mathcal{A} =\{S_{\lambda_1},S_{\lambda_2},S_{\lambda_3},\dots\}$ be an infinite subset of $\{S_n:n\in\mathbf{N}\}$, where we assume without loss of generality that $\lambda_j<\lambda_k$ whenever $j<k$. We demonstrate that 
$$\bigcup_{r=1}^{\infty}S_{\lambda_r} = \mathbf{N}$$
Proof. Prior to addressing the main claim, we first prove that result that $\lambda_{n}\ge n,\forall n\in\mathbf{N}$, by recourse to Mathematical Induction. The base case is trivial, so we assume that $\lambda_k\ge k$ for some arbitrary $k\in\mathbf{N}$, thus $\lambda_k+1\ge k+1$, but $\lambda_{k+1}>\lambda_k$, implying $\lambda_{k+1}\ge \lambda_k+1$, so $\lambda_{k+1}\ge k+1$, completing the induction. Now let $n\in\mathbf{N}$, obviously $n\in S_n$, but since $n\leq\lambda_n$, it follows that $S_n\subseteq S_{\lambda_n}$, consequently $n\in S_{\lambda_n}$ and thus $\mathbf{N}\subseteq\bigcup_{r=1}^{\infty}S_{\lambda_r}$, the converse of which is trivial.
$\blacksquare$


The idea is OK, but the proof should be set up for arbitary unions, and then the union is not always $\mathbb{N}$, though it often is:
Let $O_i$, $ i \in I$ be an arbitary union of open sets. We can WLOG omit any $O_i = \emptyset$, because they do not contribute to the union, and if any $O_i = \mathbb{N}$, so is the union and we are done. So all $O_i$ are then of the form $O_i = \{1, \ldots, n(i)\}$ for some $n(i) \in \mathbb{N}$.
Now there are two cases when considering the set $N = \{n(i): i \in I\} \subseteq \mathbb{N}$.
a) $N$ is bounded above and thus $\max N$ exists. If this is the case, $\bigcup_i O_i = \{1,\ldots, \max N\}$, as all $O_i$ are subsets of that set (as $n(i) \le \max N$) and at least one $O_i$ actually equal to that right hand set.
b) $N$ is not bounded above and then $\bigcup_i O_i = \mathbb{N}$: let $n \in \mathbb{N}$, then for some $i$, $n(i) > n$ by unboundedness, and then $n \in O_i = \{1,\ldots, n(i) \} \subseteq \bigcup_i O_i$.
So the topology is closed under all unions, as we needed to show.
