Show that for each positive integer $n$, we can choose $x_n \in U_{n} \backslash \mathbb{N} $ such that $x_n > n$. This is a problem from Morris book of "Topology withour tears" Exercise -6.1(12).
Let $X$ be the set $(\mathbb{R} \backslash \mathbb{N})\cup \{1 \} $. Define a function $f: \mathbb{R} \rightarrow X$ by 
$$
 f(x) =
  \begin{cases}
    x       & \quad \text{if } x \in \mathbb{R} \backslash \mathbb{N}  \\
    1  & \quad \text{if } x \in \mathbb{N}
  \end{cases}
$$
Further, define a topology on $\tau$ on $X$ by 
$$
\tau = \{ U : U \subseteq X \text{ and } f^{-1}(U) \text{ is open in the euclidean topology on } \mathbb{R} \}
$$
Now, you can easily prove that Every open neighbourhood of 1 in $(X; \tau )$ is of the form $(U \backslash \mathbb{N})\cup \{1 \} $ , where
$U$ is open in $\mathbb{R}$ .
In order to prove that $(X; \tau )$ is not  first countable, it starts with the argument that Suppose $(U_{1} \backslash \mathbb{N})\cup \{1 \} ; \; (U_{2} \backslash \mathbb{N})\cup \{1 \} ; \; \cdots ; (U_{n} \backslash \mathbb{N})\cup \{1 \} ; \; \cdots$ is a countable
base at 1. 
Doubt:
Show that for each positive integer $n$, we can choose $x_n \in U_{n} \backslash \mathbb{N} $ such that $x_n > n$. 
I could prove this result by assuming $U_{i}=(0,i+1) \backslash \{ 2,3,\cdots i-1 \} $ but I think this doesn't make sense if I take all my open sets inside ,say, $(0,2)$
 A: Your characterization of the open subsets of $X$ is a little bit off. To see this, note that if $A \subseteq X$ contains $1$, then $\mathbb{N} \subseteq f^{-1} (A)$, and if $f^{-1} (A)$ is additionally open in the Euclidean topology, then it includes an open interval around each natural number.
More precisely, the open neighborhoods of $1$ in $X$ look like $U \setminus ( \mathbb{N} \setminus \{ 1 \} )$, where $U$ is open in the Euclidean topology and $\mathbb{N} \subseteq U$. Putting this into a form we can work with a little bit easier, the (basic) open neighborhoods of $1$ in $X$ are of the form  $$( \textstyle{\bigcup_{n \in \mathbb{N}}} ( n-\varepsilon_n , n+\varepsilon_n) ) \setminus ( \mathbb{N} \setminus \{ 1 \} )$$
where $\varepsilon_n > 0$ for each $n \in \mathbb{N}$.  We can think about this as around each natural number $n$ you have an open neighbourhood of $n$, minus $n$ itself. (Of course, this isn't exactly correct if $\varepsilon_n > 1$, but we can avoid talking about those open sets, if we want.)
To go from here to showing that $X$ is not first countable is not too difficult. Suppose that $\{ U_k : k \in \mathbb{N} \}$ is a family of open neighborhoods of $1$ in $X$. Without loss of generality we may assume that each $U_k$ is of the form described above: $$U_k = \left( \textstyle{\bigcup_{n \in \mathbb{N}}} ( n-\varepsilon_{k,n} , n+\varepsilon_{k,n}) \right) \setminus ( \mathbb{N} \setminus \{ 1 \} )$$ where each $\varepsilon_{k,n}$ is strictly positive. Even more, we may assume that $\varepsilon_{k,n} \leq \frac{1}{2}$, so that the open intervals making up a $U_k$ do not intersect. (Making the open sets smaller only makes it more likely that they form a basis for the open neighborhoods of $1$.)
For $n \in \mathbb{N}$ define $\delta_n = \frac{\varepsilon_{n,n}}{2}$, and now set $$V = \left( \textstyle{\bigcup_{n \in \mathbb{N}}} ( n-\delta_n , n+\delta_n) \right) \setminus ( \mathbb{N} \setminus \{ 1 \} ).$$ Then $V$ is an open neighborhood of $1$, but no $U_k$ is a subset of $V$, since, for example, $k + \delta_k$ belongs to $U_k$, but not to $V$.
A: First I’m not sure you’ve got the clearest idea of the topology of $X$ and I think that’s hurting your reasoning so let’s start by talking about what the open sets of X look like. Suppose we have $1\ne A\subset X.$ Then $\Bbb N$ does not meet (ie has empty intersection with) $f^{-1}(A),$ otherwise we would have $1\in A.$ and so as $f$ is just the identity on $\Bbb R-\Bbb N$ we see that $f^{-1} A = A$. Or in other words, any open set of $\Bbb R$ which does not contain a natural number is open when viewed as a set in $X.$
Now let $1\in A\subset X.$ so $f^{-1}(A)\supset f^{-1}(1) = \Bbb N$. But we know that any open set of $\Bbb R$ which contains $x$ must also contain an open neighbourhood (which in $\Bbb R$ can’t be a singleton) of $x$ and so for $A$ to be open, $f^{-1}(A)$ must be a union of open neighbourhoods of each natural number, and we just have $A = f^{-1}(A) -\Bbb N.$ That is, we couldn’t have $A=(0.5,1.5)$ as $1\in f(A)$ so $f^{-1}(f(A)) = (0.5,1.5)\cup\Bbb N$ which is not open, so every open neighbourhood of 1 in $X$ is of the form $U-\Bbb N$ where $U\subset\Bbb R$ is open and $\Bbb N\subset U.$
Once you realise this it is straightforward to prove that $$\forall n\quad\exists x_n\in U_n\quad x_n>n.$$
Fix $n,$ and you already know $n\in U_n$ and as it is open in $\Bbb R$ there must also be some $n<x_n\in U_n$ with e.g. $x_n<n+1$ so $x_n\not\in\Bbb N$ and so $f(x_n)=x_n.$
Then there’s a diagonal argument to construct an (open) neighbourhood $V$ of 1 which does not contain any $U_i\text:$ Just enforce that $n<x_n<n+\frac12$ and set $$V=\bigcup_n \left(n-\frac12,\frac{n+x_n}2\right).$$
Then $U_n\not\subset V$ as $V\not\ni x_n\in U_n.$
