Let $X$ be any set, and let $d_{disc}$ be the discrete metric on $X$. Let $(x^{(n)})_{n=m}^{\infty}$ be a sequence of points in $X$, and let $x$ be a point in $X$. Then $(x^{(n)})_{n=m}^{\infty}$ converges to $x$ with respect to the discrete metric $d_{disc}$ iff there exists an $N\geq m$ such that $x^{(n)} = x$ for all $n\geq N$.
My solution
According to the definition of convergence, for every $\varepsilon > 0$, there exists an $N\geq m$ such that \begin{align*} n\geq N \Rightarrow d(x^{(n)},x) \leq \varepsilon \end{align*} Consequently, if we choose $\varepsilon = 1/2$, we conclude there is an $N_{1/2}\geq m$ such that \begin{align*} n\geq N_{1/2}\Rightarrow d(x^{(n)},x)\leq 1/2 \end{align*}
Since $d(a,b) = 1$ when $a\neq b$ and $d(a,b) = 0$ otherwise, we conclude that $d(x^{(n)},x) = 0$ for $n\geq N_{1/2}$.
In other words, $x^{(n)} = x$ for every $n\geq N_{1/2}$ as claimed, and the results follows.
Could someone please verify if is it correct?