Proof

Assume, the sequence of functions $f_n:D\to\mathbb F$ is uniformly convergent to a function $f:D\to\mathbb F.$
The real sequence of supremum norms $(||f_n-f||_\infty)_{n\to\infty}$ is therefore convergent.
By the lemma convergent sequences are Cauchy sequences, the sequence $(||(f_n-f)||_\infty)_{n\to\infty}$ is a Cauchy sequence.
By definition, this means for every $\epsilon > 0$ there is an index $N$ such that the supremum norm $||(f_n-f)-(f_m-f)||_\infty=||f_n-f_m||_\infty < \epsilon$ for all $n,m\ge N.$

Assume, for every $\epsilon > 0$ there is an index $N$ such that the supremum norm $||f_n-f_m||_\infty < \epsilon$ for all $n,m\ge N.$
This means that $|f_n(x)-f_m(x)|<\epsilon$ for all $x\in D$ and all $n,m\ge N.$
By definition of Cauchy sequences, this means that $(f_n(x))_{n\in\mathbb N}$ is a Cauchy sequence for all $x\in D.$
By the completeness principle, this means that there is a limit $y:=\lim_{n\to\infty}f_n(x)$ for all $x\in D.$
In other words, there is a function $f:D\to\mathbb F$ to which the sequence of functions $f_n:D\to\mathbb F$ is uniformly convergent.
∎

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Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition