It is not possible to have a smooth metric. Non-smoothness at the diagonal is inevitable.
Suppose you had such a metric on a manifold $M$. Take a smooth curve $\gamma:(-1,1)\to M$ with $\gamma'(x)\neq0$. Consider the function $f(t,s)=d(\gamma(t),\gamma(s))$, which is now smooth.
Since $$ \begin{split} 0 &= \frac{d}{dt}0 \\&= \frac{d}{dt}f(t,t) \\&= \partial_1f(t,t)+\partial_2f(t,t), \end{split} $$ we have $\partial_1f=-\partial_2f$ on the diagonal. But since $f(t,s)=f(s,t)$, we also have $\partial_1f=\partial_2f$ on the diagonal. Thus $\partial_1f(t,t)=\partial_2f(t,t)=0$.
This implies that $f(t,s)\leq C(t-s)^2$ for some constant $C$, for all $t,s\in[-1/2,1/2]$. Pick any number $a\in[0,\frac12]$ and a large integer $N$ and observe that by triangle inequality $$ \begin{split} d(0,a) &\leq d(0,\frac{a}{N})+d(\frac{a}{N},\frac{2a}{N})+\dots+d(\frac{(N-1)a}{N},a) \\&\leq NC\left(\frac aN\right)^2 \\&= \frac{Ca^2}{N}. \end{split} $$ This holds for any $N$, so in fact $d(0,a)=0$. This implies that as long as $x,y\in M$ are in the same connected component, they must satisfy $d(x,y)=0$.