Let $\sigma: [0,1] \to (X,d)$ be a continuous and rectifiable curve. Then $\sigma: [0,1] \to (X,d_i)$ is also a continuous and rectifiable curve and $L_{d_i}(\sigma) = L_{d}(\sigma)$.
Let $\varepsilon \gt 0$. Since $[0,1]$ is compact, $\sigma: [0,1] \to (X,d)$ is uniformly continuous, so there is $\delta \gt 0$ such that $|t-t'| \lt \delta$ implies $d(\sigma(t),\sigma(t')) \leq \varepsilon/2$. On the other hand, as $\sigma$ is rectifiable, $L(\sigma)$ is finite, so we can find a partition $0 = t_0 \lt t_1 \lt \cdots \lt t_{n-1} \lt t_{n} = 1$ such that $ L(\sigma) - \varepsilon /2 \lt \sum_{k=1}^n d(\sigma(t_k),\sigma(t_{k-1})) \leq L(\sigma). $ The triangle inequality yields that refining the partition only increases the sum in this estimate, so we may assume that $t_k \lt t_{k+1} \lt t_{k} + \delta$ to begin with. But $L(\sigma) = \sum_{k=1}^n L(\sigma|_{[t_{k-1},t_k]})$, so $ 0 \leq \sum_{k=1}^n \left( L(\sigma|_{[t_{k-1},t_k]}) - d(\sigma(t_k),\sigma(t_{k-1}))\right) \leq \varepsilon/2. $ Each summand in the last sum is non-negative, in particular $ L(\sigma|_{[t_{k-1},t_k]}) \leq d(\sigma(t_k),\sigma(t_{k-1})) + \varepsilon /2 \lt \varepsilon. $ This tells us two things:
- $t \mapsto L(\sigma|_{[0,t]})$ is continuous (and non-decreasing)
- $d_i(\sigma(t_{k}),\sigma(t_{k-1}))\lt \varepsilon$, in particular $\sigma:[0,1]\to (X,d_i)$ is uniformly continuous.
Finally, $L_{d_i}(\sigma) = \sup_{0=t_0 \lt \cdots \lt t_n=1} \sum_{k=1}^n d_i(\sigma(t_k),\sigma(t_{k-1}) \leq \sup_{0=t_0 \lt \cdots \lt t_n=1} \sum_{k=1}^n L_d(\sigma|_{[t_{k-1},t_k]}) = L_d(\sigma) \leq L_{d_i}(\sigma),$ where the first inequality follows from the definition of $d_i$ and the second inequality from the fact that $d \leq d_i$.
To finish your question up, if $d_i(x,y) \lt \infty$ for all $x,y \in X$ we have $(d_i)_i = d_i$ by the above:
For every $\varepsilon \gt 0$ we may find a continuous and rectifiable curve $\sigma: [0,1] \to (X,d)$ such that $L_{d}(\sigma) \leq d_i(x,y) + \varepsilon$ but then $L_{d}(\sigma) = L_{d_i}(\sigma)$ gives that $(d_i)_i(x,y) \leq d_i(x,y)+\varepsilon$. Thus, $d_i = (d_i)_i$.