Let $X$ be a metric space with metric $d$.Then $d:X \times X\rightarrow \mathbb R$ is continuous?
Please Check the following proof-
We'll try to show this via sequential crieria of continuity.Let$(x,y)\in X\times X$ be any arbitrary point.We need to show that $d$ is continuous at $(x,y)$.
Let $<(x_n,y_n)>$ be a sequence in $X\times X$ converges to $(x,y)\in X\times X$.Now we'll show that $<d(x_n,y_n)>$ converges to $d(x,y)$.
Since,$\lim_{n\rightarrow {\infty}}(x_n,y_n)=(x,y)$.Therefore,$\lim_{n\rightarrow {\infty}}x_n=x$ &$\lim_{n\rightarrow {\infty}}y_n=y\implies d(x_n,x)<\frac{\epsilon}{2}\forall n>m_1$ and $d(y_n,y)<\frac{\epsilon}{2}\forall n>m_2$.
Now,$d(x_n,y_n)\leq d(x_n,x)+d(x,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n) \implies d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y,y_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
i.e$d(x_n,y_n)-d(x,y)<\epsilon$ or $\lim_{n\rightarrow{\infty}}d(x_n,y_n)=d(x,y)$ .
Hence,$\lim_{n\rightarrow{\infty}}(x_n,y_n)=(x,y)\implies \lim_{n\rightarrow{\infty}}d(x_n,y_n)=d(x,y)$. Hence,$d$ is continuous on $X$.