In general, the answer is negative. I found an abstract of an article “A new class of spaces with all finite powers Lindelöf” by Natasha May, Santi Spadaro, and Paul Szeptycki. (Topology Appl. 170, 104-118 (2014)):
An $\omega$-cover of a set $X$ is a cover with the following property: every finite subset of $X$ is contained in an element of the cover. An $\iota$-cover is an $\omega$-cover with the additional property that disjoint finite sets are separated by members of the cover (meaning that if $F, G \in [X]^{<\omega}$ are disjoint and $\mathcal{U}$ is an $\iota$-cover then there is $U \in \mathcal{U}$ such that $F \subseteq U$ and $G \cap U = \emptyset$).
A topological space $X$ is called an $\epsilon$-space if every finite power of $X$ is Lindelöf; equivalently, $X$ is an $\epsilon$-space if every open $\omega$-cover has a countable $\omega$-subcover. The authors remark that every space with a countable network is an $\epsilon$-space, and they show that every Hausdorff space with a countable network has the property that every open $\iota$-cover has a countable refinement that is also an $\iota$-cover. Motivated by this, they call a space with the latter property an $\iota$-space.
In the paper under review, the authors investigate a number of issues related to $\epsilon$-spaces and $\iota$-spaces. A ZFC example of a regular $\iota$-space with no countable network is provided. The authors show that the property of being an $\epsilon$-space and the property of being an $\iota$-space are equivalent for regular spaces with a $G_\delta$ diagonal. A consistent example of a hereditarily $\epsilon$-space whose square is not hereditarily Lindelöf is constructed, as well as an absolute (meaning, ZFC) example of a non $D$-space that has a countable open $\iota$-cover. The paper ends with the following questions: is every $\iota$-space a $D$-space? Is every hereditarily $\iota$-space a $D$-space?
So, the answer is positive iff every finite power of the space $X$ is Lindelöf. In particlular, the answer is positive when $X$ has a countable network or is $\sigma$-compact. But there is a classical example of a Lindelöf space which square is not Lindelöf. It is so-called Sorgenfrey arrow $\Bbb L$, that is the real line endowed with the Sorgenfrey topology generated by the base consisting of half-intervals $[a,b)$, $a<b$. In is well known that the space $\Bbb L$ is Lindelöf, but its square contains a closed discrete space $\{(x,-x):x\in\Bbb R\}$ of cardinality $\frak c$.