Let $(X,d)$ be a metric space, and let $f_{n}$ be a sequence of bounded real-valued continuous functions on $X$ such that the series $\sum_{n=1}^{\infty}\|f_{n}\|$ is convergent. Then the series $\sum_{n=1}^{\infty}f_{n}$ converges uniformly to some function $f$ on $X$, and that function $f$ is also continuous.
MY ATTEMPT
We want to prove that for every $\varepsilon > 0$, there is a natural $N\geq 0$ such that for every $x\in X$ one has that \begin{align*} n\geq N \Rightarrow\left|\sum_{j=1}^{n}f_{j}(x) - f(x)\right| < \varepsilon \end{align*}
This is the same as proving that $\displaystyle s_{n} = \displaystyle\sum_{j=1}^{n}f_{j}(x)$ is Cauchy for every $x\in X$, since $\textbf{R}$ is complete.
Suppose that $\sum\|f_{n}\|$ converges. Then it is a Cauchy sequence. This means that for every $\varepsilon > 0$, there exists a $N\geq 0$ such that \begin{align*} n\geq m \geq N \Rightarrow \sum_{j=m+1}^{n}\|f_{j}\| < \varepsilon \end{align*}
But we do also know that \begin{align*} \|f_{n}\| = \sup_{x\in X}|f_{n}(x)| \end{align*}
Consequently, we have that for every $\varepsilon > 0$, there exists a natural number $N\geq 0$ such that for every $x\in X$ \begin{align*} n\geq m \geq N \Rightarrow \left|\sum_{j=m+1}^{n}f_{j}(x)\right| \leq \sum_{j=m+1}^{n}|f_{j}(x)| \leq \sum_{j=m+1}^{n}\|f_{j}\| < \varepsilon \end{align*}
whence we conclude that $s_{n}$ is Cauchy for every $x\in X$, and we are done.
My concerns
I am new to this and I am a little bit unsure about the Cauchy condition part.
Could someone tell me if I am proving it correctly?