I read that given $X = (\mathbb{R} \setminus \mathbb{N}) \cup \{1\}$, we can define a continuous function $f : \mathbb{R} \rightarrow X$, such that:
$$ f (x)=\begin{cases} x, & \text{if}\ x \in \mathbb{R} \setminus \mathbb{N} \\ 1, & \text{if}\ x \in \mathbb{N} \end{cases} $$
We also define a topology $\tau$ on $X$ as follows (such that $X$ is not first-countable):
$$ \tau = \{U : U \subseteq X \text{ and } f^{-1}(U) \text{ is open in the euclidian topology on } \mathbb{R} \} $$
Why is $f$ continuous given that there are discontinuities at $x \in \mathbb{N}$? and how is $X$ not first countable?
Given you take $\tau$ to be the topology of $X$, $f$ is continuous by definition. $\tau$ is the smallest topology for which $f$ is countable. If $X$ is equipped with the induced topology then $f$ is not continuous as $f^{-1}(\frac{1}{2},\frac{3}{2})=(\frac{1}{2},\frac{3}{2})\cup\mathbb{N}$ which is not an open set in $\mathbb{R}$.
To see that $X$ is not first countable, suppose that $(U_{n})$ is a neighbourhood base of $1$. Clearly for all $n\in\mathbb{N}$ we can find a function $r_{n}:\mathbb{N}\rightarrow(0,1)$ such that $(i-r_{n}(i),i+r_{n}(i))\subset U_{n}$.
Now we define a function $s:\mathbb{N}\rightarrow(0,\frac{1}{2})$ by $$s(i)=\frac{1}{2}\min_{1\leq k\leq i}r_{n}(i)$$ and we define $$U=\bigcup^{\infty}_{i=1}(i-s(i),i+s(i)).$$ Note that $U$ is a neighbourhood of $1$, but clearly for all $n$ we have $U_{n}\not\subset U$. Which contradicts that $(U_{n})$ is a neighbourhood base.