Suppose that $(x_n) \equiv (y_n)$ and that $(y_n) \equiv (z_n)$. The goal is to show that $(x_n) \equiv (z_n)$, which is equivalent to showing that the "merged" sequence $(x_1, z_1, x_2, z_2, \dotsc)$ is Cauchy. To do this, for each $\varepsilon > 0$, it is sufficient to find some $N$ so large that $m,n>N$ implies that $$ |x_m - x_n| < \varepsilon, \qquad |z_m - x_n| < \varepsilon, \qquad \text{and} \qquad |z_m - z_n| < \varepsilon. \tag{*}\label{whatwewant}$$ Note that this condition is sufficient, but it is a slight modification of the usual statement of the Cauchy criterion.
Since $(x_n)$ and $(z_n)$ are Cauchy, there are $N_1$ and $N_2$ such that $m,n > N_1$ implies that $$ |x_m - x_n| < \varepsilon, $$ and $m,n > N_2$ implies that $$ |z_m - z_n| < \varepsilon, $$ so those two bounds from (\ref{whatwewant}) are taken care. All that is left is find an $N_3$ large enough to give the middle bound. By the assumption that $(x_n) \equiv (y_n)$, there is an $M_1$ such that $m,n>M_1$ implies that $$ |x_n - y_m| < \frac{\varepsilon}{2}. $$ There are other inequalities which are also satisfied, e.g. $|x_m - x_n| < \varepsilon/2$ for this $M_1$, but these other inequalities are not particularly relevant to the argument here. Similarly, since $(y_n) \equiv (z_n)$, there is an $M_2$ such that $m,n > M_2$ implies that $$ |y_m - z_n| < \frac{\varepsilon}{2}. $$ Choose $N_3 = \max\{M_1, M_2\}$ and note that if $m, n > N_3$, then $$ |x_m - z_n| = |x_m - y_m + y_m - z_n| \le |x_m - y_m| + |y_m - z_n| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. $$ Taking $N = \max\{N_1, N_2, N_3\}$, if $m, n > N$ then all three of the bounds in (\ref{whatwewant}) are satisfied.
NB: After a fair amount of time searching the site, I cannot find a duplicate question—I find this somewhat surprising, as this seems like a kind of natural question to ask if Cauchy sequences are going to be introduced in this way. My conclusion is that this is somewhat idiosyncratic presentation—the usual approach, I think, is to define a null sequence to be a Cauchy sequence $(\xi_n)$ such that $\xi_n \to 0$. Then one shows constructs a complete metric space by taking the entire set of Cauchy sequences and modding out by the set of null sequences. This will, by all the usual algebraic nonsense about what it means to "mod out" one space by another, automatically create equivalence classes of Cauchy sequences (where equivalence is up to a null sequence, i.e. the difference between the two sequences converges to zero).
One might want to show that the equivalence classes constructed in this manner are the same as those constructed by "merging" sequences. Given two Cauchy sequences $(x_n)$ and $(y_n)$, the two conditions are
- $(x_n - y_n)$ is a null sequence, and
- $(x_1, y_1, x_2, y_2, \dotsc)$ is a Cauchy sequence.
That is the second implies the first is not too hard—if a sequence is Cauchy, then the difference between consecutive terms goes to zero (almost by definition). The other direction requires a bit more thinking, but can probably be done in two or three sentences (it comes down to the fact that $(x_n)$, $(y_n)$, and $(x_n - y_n)$ are all Cauchy, plus maybe a little triangle inequality).
