Given a sequence $(f_n)$ of functions $f_n:X \to [0,\infty]$ where $f_n = \chi_{[0,n]}$, it is easy to see that $(f_n)$ converges to $\chi_{[0,\infty)}$ pointwise. However, I believe that $(f_n)$ does not converge to $\chi_{[0,\infty)}$ uniformly for the following reason.
Choose $x = n+1$, then $\sup_{x}|\chi_{[0,\infty)}(x) - \chi_{[0,n]}(x)| \geq |\chi_{[0,\infty)}(n+1) - \chi_{[0,n]}(n+1)| = 1$. Therefore $1 \leq \lim_{n \to \infty}\sup_{x}|\chi_{[0,\infty)}(x) - \chi_{[0,n]}(x)| \neq 0$.
Is my reasoning logical? Any help will be greatly appreciated.
It is fine. However, the expression$$1\leqslant\lim_{n\to\infty}\sup_x\bigl\lvert\chi_{[0,\infty)}(x)-\chi_{[0,n]}(x)\bigr\rvert\neq0$$assumes that the limit $\lim_{n\to\infty}\sup_x\bigl\lvert\chi_{[0,\infty)}(x)-\chi_{[0,n]}(x)\bigr\rvert$ exists. It would be better if you either observe that you always have $\sup_x\bigl\lvert\chi_{[0,\infty)}(x)-\chi_{[0,n]}(x)\bigr\rvert=1$, in which case it is clear that the limit exists, or to note that, since $\sup_x\bigl\lvert\chi_{[0,\infty)}(x)-\chi_{[0,n]}(x)\bigr\rvert\geqslant1$, then the limit $\lim_{n\to\infty}\sup_x\bigl\lvert\chi_{[0,\infty)}(x)-\chi_{[0,n]}(x)\bigr\rvert$ either doesn't exist or, if it does, it is at least $1$.