Suppose you have $E=\mathbb{R}^{\mathbb{N}}$. $\forall f,g \in E$ we have $$d(f,g)= \sum^{\infty}_{k=0}\frac{1}{2^k} \min(|f(k)-g(k)|,1)$$ Show that every Cauchy sequence in $(E,d)$ converges.
Now, I am unable to prove that because simply because if you take $(u_n),(v_n) \in E$, I don't know what for an $m \in \mathbb{N}$, what $d(u_m,v_m)$ is equal to. Seems like there is no way to calculate the distance for a particular natural number. Can someone help me how I can interpret this distance?
It makes no sense to speak of the "distance" $d(f_n,g_n)$ because it isn't defined. The distance function $d$ takes as input two sequences, not two sequence elements (real numbers). You are being asked to take a Cauchy sequence $(f_n)$ in $E$, i.e., a sequence of sequences, and to show that this sequence converges in $E$.
Hint: let $(f_n)$ be a Cauchy sequence in $E$. This means in particular that each $f_n = (f_n(k))$ is itself a sequence of real numbers. The Cauchy criterion means that for any $\epsilon > 0$, $$\sum_{k=0}^\infty \frac{1}{2^k} \min(\lvert f_n(k) - f_m(k)\rvert,1) < \epsilon$$ for large enough $n,m$. This implies that for every fixed $k = 0,1,2,\ldots$ the real sequence $a_k$ given by $a_k(n) = f_n(k)$ is Cauchy (why?). Hence by completeness of $\mathbb{R}$, each real sequence $a_k$ converges to some number $c_k$, $a_k(n) \xrightarrow{n \to \infty} c_k \in \mathbb{R}$. Finally, show that $(f_n(k)) \xrightarrow{n \to \infty} (c_k) \in E$.