The answer is **yes** with $k = 2d + 1$. (Can this be improved?)

It is a by-product of the so-called Reidemester-Schreier method [1, Proposition II.4.1].

> **Claim**. Let $G$ be a group and let $H$ be a subgroup of $G$ with index $d$ in $G$. Let $S$ be a generating subset of $G$. Then $H$ can be generated by elements of length at most $k = 2d + 1$ with respect to $S$.

> *Proof.* Using the notation of [1, Proposition II.4.1], we consider a group presentation $F/N$ of $G$ where $F$ is a free group with basis $X$ which maps to $S$ via the quotient map $F \twoheadrightarrow F/N$, $\tilde{H} \subseteq F$ the preimage of $H$ in $F$ and $T$ a Schreier transversal of $\tilde{H}$ in $F$. Then $H$ is generated by elements of the form $\gamma(x, t) = tx (\overline{tx})^{-1}$ with $x \in X$, where $\overline{w} \in T$ is defined through $\tilde{H}w = \tilde{H}\overline{w}$ for $w \in F$. The result follows since every word in $T$ has length at most $d$ with respect to $X$.

**Note.** A Schreier transversal of a subgroup $H$ of $F$, free with basis $X$, is a subset $T$ of $F$ such that the cosets $Ht$ are distinct for $t \in T$, the union of the cosets $Ht$ is $F$ and every initial segment of an element $t \in T$, as a word over $X^{\pm 1}$, is in $T$.

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[1] R. Lyndon and P. Schupp, "Combinatorial Group Theory", 1977.