We have that the supremum metric on $A \times A$ is defined as: :$\displaystyle \forall x, y \in A: d \left({x, y}\right) := \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert$ where $x = \left\langle{x_i}\right\rangle$ and $y = \left\langle{y_i}\right\rangle$ are bounded real sequences. So: :$\exists K, L \in \R: \left\vert{x_n}\right\vert \le K, \left\vert{y_n}\right\vert \le L$ for all $n \in \N$. First note that we have: {{begin-eqn}} {{eqn | l = \left\vert{x_n - y_n}\right\vert       | r = \left\vert{x_n + \left({- y_n}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{x_n}\right\vert + \left\vert{\left({- y_n}\right)}\right\vert       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \left\vert{x_n}\right\vert + \left\vert{y_n}\right\vert       | c = {{Defof|Absolute Value}} }} {{eqn | o = \le       | r = K + L       | c =  }} {{end-eqn}} and so the {{RHS}} exists. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = d \left({x, x}\right)       | r = \sup_{n \mathop \in \N} \left\vert{x_n - x_n}\right\vert       | c = Definition of $d$ }} {{eqn | r = \sup_{n \mathop \in \N} \left\vert{0}\right\vert       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d$. {{qed|lemma}} === Proof of $M2$ === Let $x, y, z \in A$. Let $n \in \N$. {{begin-eqn}} {{eqn | l = n       | o = \in       | r = \N       | c =  }} {{eqn | ll= \implies       | l = \left\vert{x_n - z_n}\right\vert       | o = \le       | r = \left\vert{x_n - y_n}\right\vert + \left\vert{y_n - z_n}\right\vert       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert + \sup_{n \mathop \in \N} \left\vert{y_n - z_n}\right\vert       | c = {{Defof|Supremum of Real Sequence}} }} {{eqn | r = d \left({x, y}\right) + d \left({y, z}\right)       | c = Definition of $d$ }} {{end-eqn}} Thus $d \left({x, y}\right) + d \left({y, z}\right)$ is an upper bound for: :$S := \left\{ {\left\vert{x_n - z_n}\right\vert: n \in \N}\right\}$ So: :$d \left({x, y}\right) + d \left({y, z}\right) \ge \sup S = d \left({x, z}\right)$ So axiom $M2$ holds for $d$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d \left({x, y}\right)       | r = \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert       | c = Definition of $d$ }} {{eqn | r = \sup_{n \mathop \in \N} \left\vert{y_n - x_n}\right\vert       | c = {{Defof|Absolute Value}} }} {{eqn | r = d \left({y, x}\right)       | c = Definition of $d$ }} {{end-eqn}} So axiom $M3$ holds for $d$. {{qed|lemma}} === Proof of $M4$ === As $d$ is the supremum of the absolute value of the difference of termselements of the sequences $\left\langle{x_i}\right\rangle$ and $\left\langle{y_i}\right\rangle$: :$\forall x, y \in A: d \left({x, y}\right) \ge 0$ Suppose $x, y \in A: d \left({x, y}\right) = 0$. Then: {{begin-eqn}} {{eqn | l = d \left({x, y}\right)       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert       | r = 0       | c = Definition of $d$ }} {{eqn | ll= \implies       | lo= \forall n \mathop \in \N:       | l = x_n        | r = y_n       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \implies       | l = x       | r = y       | c = Equality of Mappings }} {{end-eqn}} So axiom $M4$ holds for $d$. {{qed}}