Find an example of a sequence $\{f_n\}_{n \in \Bbb Z_+}$ of real-valued functions defined on any metric space $X$ such that the series $\sum_{n \in \Bbb Z_+} f_n$ converges pointwise to $f$, each $f_n$ is uniformly continuous but $f$ is not.
My Attempt
Let $X = ]1, 2]$ with the usual Euclidean Metric, then define: $$f_n(x) = \left( \frac{1}{x} \right)^{n}$$ which is uniformly continuous for every $n \in \Bbb Z_+$ because it is actually Lipschitz on ]1, 2] (it suffices to check that the derivative is bounded above). Now, it is easy to verify that: $$f(x) = \sum_{n \in \Bbb Z_+}f_n(x) = \frac{x}{x-1}, \forall x \in ]1, 2]$$ however, near $x = 1$, $f$ explodes to infinity more rapidly than a linear function, hence it can not be uniformly continuous (one can also check it by contradiction).
Is this argument good enough?
As always any answer or hint is welcome and let me know if I can explain myself clearer!