The answer is Yes for complex manifolds of dimension one.
Indeed for any open subset $U\subset X$ the long exact cohomology sequence associated to the exponential sequence you mention yields the fragment $$\cdots\to H^i(U,\mathcal O_X)\to H^i(U,\mathcal O_X^*)\to H^{i+1}(U,\mathbb Z)\to \cdots$$ Now $U$, like any non-compact Riemann surface, is Stein and thus $H^i(U,\mathcal O_X)=0$ for $i\geq 1$.
And $ H^{i+1}(U,\mathbb Z)=0$ for $i\geq 1$: for dimension reason if $i\geq 2$ and because $U$ is non-compact $i=1$.
In conclusion $H^i(U,\mathcal O_X^*)=0$ ($i\geq 1$) for $U$ non-compact of dimension one, which of course shows that any covering is acyclic for $\mathcal O^*_X$.

Note carefully
a) On a non-compact Riemann surface all holomorphic vector bundles, of whatever rank, are trivial !
This amazing result is Theorem 30.4, page 229, of Forster's wonderful book Lectures on Riemann Surfaces.

b) The acyclicity result in the answer is completely false for a smooth algebraic curve $Y$ of dimension one over $\mathbb C$ of dimension one, for example for the algebraic curve underlying a Riemann surface.
Indeed for any open (in the Zariski topology!) subset $V\subset Y $, the group $H^1(V, \mathcal O_{Y,\operatorname {alg}}^*)=\operatorname {Pic}_{alg}(V)$ is non denumerable. This means that there are more than denumerably many non-isomorphic algebraic line bundles on $V$ which are all holomorphically trivial!

The answer is Yes for complex manifolds of dimension one.
Indeed for any open subset $U\subset X$ the long exact cohomology sequence associated to the exponential sequence you mention yields the fragment $$\cdots\to H^i(U,\mathcal O_X)\to H^i(U,\mathcal O_X^*)\to H^{i+1}(U,\mathbb Z)\to \cdots$$ Now $U$, like any non-compact Riemann surface, is Stein and thus $H^i(U,\mathcal O_X)=0$ for $i\geq 1$.
And $ H^{i+1}(U,\mathbb Z)=0$ for $i\geq 1$: for dimension reason if $i\geq 2$ and because $U$ is non-compact $i=1$.
In conclusion $H^i(U,\mathcal O_X^*)=0$ ($i\geq 1$) for $U$ non-compact of dimension one, which of course shows that any covering is acyclic for $\mathcal O^*_X$.

Remarks
a) On a non-compact Riemann surface all holomorphic vector bundles, of whatever rank, are trivial !
This amazing result is Theorem 30.4, page 229, of Forster's wonderful book Lectures on Riemann Surfaces.

b) The acyclicity result in the answer is completely false for a smooth algebraic curve $Y$ of dimension one over $\mathbb C$ of dimension one, for example for the algebraic curve underlying a Riemann surface.
Indeed for any open (in the Zariski topology!) subset $V\subset Y $, the group $H^1(V, \mathcal O_{Y,\operatorname {alg}}^*)=\operatorname {Pic}_{alg}(V)$ is non denumerable. This means that there are more than denumerably many non-isomorphic algebraic line bundles on $V$ which are all holomorphically trivial!

c) See also this answer.

The answer is NO!
For example, let $X=S\setminus\{s\}$ be the complex manifold obtained from a compact Riemann surface $S$ of genus $\geq 1$ by removing an arbitrary point $s\in S$ .
Then every non-empty open subset $U\subset X$ has non denumerable Picard group $\operatorname {Pic}(U)=H^1(U, \mathcal O^*)$ , which proves that the acyclic cover you require does not exist.
(By the way, note that $X$, like all non compact Riemann surfaces, is Stein)

The answer is Yes for complex manifolds of dimension one.
Indeed for any open subset $U\subset X$ the long exact cohomology sequence associated to the exponential sequence you mention yields the fragment $$\cdots\to H^i(U,\mathcal O_X)\to H^i(U,\mathcal O_X^*)\to H^{i+1}(U,\mathbb Z)\to \cdots$$ Now $U$, like any non-compact Riemann surface, is Stein and thus $H^i(U,\mathcal O_X)=0$ for $i\geq 1$.
And $ H^{i+1}(U,\mathbb Z)=0$ for $i\geq 1$: for dimension reason if $i\geq 2$ and because $U$ is non-compact $i=1$.
In conclusion $H^i(U,\mathcal O_X^*)=0$ ($i\geq 1$) for $U$ non-compact of dimension one, which of course shows that any covering is acyclic for $\mathcal O^*_X$.

Note carefully
a) On a non-compact Riemann surface all holomorphic vector bundles, of whatever rank, are trivial !
This amazing result is Theorem 30.4, page 229, of Forster's wonderful book Lectures on Riemann Surfaces.

b) The acyclicity result in the answer is completely false for a smooth algebraic curve $Y$ of dimension one over $\mathbb C$ of dimension one, for example for the algebraic curve underlying a Riemann surface.
Indeed for any open (in the Zariski topology!) subset $V\subset Y $, the group $H^1(V, \mathcal O_{Y,\operatorname {alg}}^*)=\operatorname {Pic}_{alg}(V)$ is non denumerable. This means that there are more than denumerably many non-isomorphic algebraic line bundles on $V$ which are all holomorphically trivial!