I am trying to prove that $d_2\left({x,y}\right)=\sum_{j=1}^{k}\left|{x_j-y_j}\right|$ is a complete metric on $\mathbb{R}^k$. Here is my reasoning for why it is, but I am unsure about its correctness.
The first three properties of a metric are omitted for brevity, where as the triangle inequality property can be shown as follows. Since $\left|{x_j-y_j}\right| \leq \left|{x_j-z_j}\right| + \left|{z_j-y_j}\right|$, summing up this inequality for $j=1,2,...,k$ gives $\sum_{j=1}^{k}\left|{x_j-y_j}\right| \leq \sum_{j=1}^{k}\left|{x_j-z_j}\right| + \sum_{j=1}^{k}\left|{z_j-y_j}\right|$. Hence, $d_2\left({x,y}\right) \leq d_2\left({x,z}\right) + d_2\left({z,y}\right)$ and is a metric on $\mathbb{R}^k$.
To show that $d_2\left({x,y}\right)$ is complete, I must show that a Cauchy sequence $\left({x_n}\right)$ converges to an element of $\mathbb{R}^k$, i.e. that $x_j^n$ for each $j=1,2,...,k$ converges to an element of $\mathbb{R}$. So I must determine whether there exists $N$ such that $n,m>N\implies \left|{x_j^n-x_j^m}\right|<\epsilon$. Since $d\left({x,y}\right)$ is a complete metric on $\mathbb{R}$, and for any $j$ we see that $\left|{x_j^n-x_j^m}\right|=d\left({x_j^n,x_j^m}\right)$ it follows that $x_j^n$ converges in $\mathbb{R}$ for each $j=1,2,...,k$ and thus $\left({x_n}\right)$ converges in $\mathbb{R}^k$.
Almost there! You're a bit confused on the exposition.
It's true that $|x_j-y_j|\leq d(x,y)$, but saying $|x_j-y_j|$ converges makes no sense, since this is not even a sequence; rather, this is the difference in the $j$-th components of $x$ and $y$. Go back to your original sequence $(x_n)$ and apply what you've just said. You want to say that the the components of the $x_n$-s are Cauchy sequences. From this, you can actually extract the components of the limit of your sequence $(x_n)$. How?