Theorem. The following are equivalent for any Hausdorff space X.

1. X is compact.

2. X^κ is Lindelöf for any cardinal κ.

3. X^ω₁ is Lindelöf.

Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if X is compact, then X^κ is compact and hence Lindelöf.

So suppose that we have a space X that is not compact, but X^ω₁ is Lindelöf. It follows that X is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets U₀ subset U₁ subset ... U_n ... with the union U{ U_n | n in ω} = X.

For each J subset ω₁ of size n, let U_J be the set {s in X^ω₁ | s(α) in U_n for each α in J}. As the size of J increases, the set U_J allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call U_J an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such U_J for all finite J subset ω₁.

This F is a cover of X^ω₁. To see this, consider any point s in X^ω₁. For each α in ω₁, there is some n with s(α) in U_n. Since ω₁ is uncountable, there must be some value of n that is repeated unboundedly often, in particular, some n occurs at least n times. Let J be the coordinates where this n appears. Thus, s is in U_J. So F is a cover.

Since X^ω₁ is Lindelöf, there must be a countable subcover F₀. Let J* be the union of all the finite J that appear in the U_J in this subcover. So J* is a countable subset of ω₁. Note that J* cannot be finite, since then the sizes of the J appearing in F₀ would be bounded and it could not cover X^ω₁. We may rearrange indices and assume without loss of generality that J*=ω is the first ω many coordinates. So F₀ is really a cover of X^ω, by ignoring the other coordinates.

But this is impossible. Define a sequence s in X^ω₁ by choosing s(n) to be outside U_n+1, and otherwise arbitrary. Note that s is in U_n in fewer than n coordinates below ω, and so s is not in any n-box with J subset ω, since any such box has n values in U_n. Thus, s is not in any set in F₀, so it is not a cover. QED

In particular, to answer the question at the end, it suffices to take any uncountable kappa.

Theorem. The following are equivalent for any Hausdorff space $X$.

1. $X$ is compact.

2. $X^\kappa$ is Lindelöf for any cardinal $\kappa$.

3. $X^{\omega_1}$ is Lindelöf.

Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if $X$ is compact, then $X^\kappa$ is compact and hence Lindelöf.

So suppose that we have a space $X$ that is not compact, but $X^{\omega_1}$ is Lindelöf. It follows that $X$ is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets $U_0 \subset U_1 \subset \dots U_n \dots$ with the union $\bigcup\lbrace U_n \; | \; n \in \omega \rbrace = X$.

For each $J \subset \omega_1$ of size $n$, let $U_J$ be the set $\lbrace s \in X^{\omega_1} \; | \; s(\alpha) \in U_n$ for each $\alpha \in J \rbrace$. As the size of $J$ increases, the set $U_J$ allows more freedom on the coordinates in $J$, but restricts more coordinates. If $J$ has size $n$, let us call $U_J$ an open $n$-box, since it restricts the sequences on $n$ coordinates. Let $F$ be the family of all such $U_J$ for all finite $J \subset \omega_1$

This $F$ is a cover of $X^{\omega_1}$. To see this, consider any point $s \in X^{\omega_1}$. For each $\alpha \in \omega_1$, there is some $n$ with $s(\alpha) \in U_n$. Since $\omega_1$ is uncountable, there must be some value of $n$ that is repeated unboundedly often, in particular, some $n$ occurs at least $n$ times. Let $J$ be the coordinates where this $n$ appears. Thus, $s$ is in $U_J$. So $F$ is a cover.

Since $X^{\omega_1}$ is Lindelöf, there must be a countable subcover $F_0$. Let $J^*$ be the union of all the finite $J$ that appear in the $U_J$ in this subcover. So $J^*$ is a countable subset of $\omega_1$. Note that $J^*$ cannot be finite, since then the sizes of the $J$ appearing in $F_0$ would be bounded and it could not cover $X^{\omega_1}$. We may rearrange indices and assume without loss of generality that $J^*=\omega$ is the first $\omega$ many coordinates. So $F_0$ is really a cover of $X^\omega$, by ignoring the other coordinates.

But this is impossible. Define a sequence $s \in X^{\omega_1}$ by choosing $s(n)$ to be outside $U_{n+1}$, and otherwise arbitrary. Note that $s$ is in $U_n$ in fewer than $n$ coordinates below $\omega$, and so $s$ is not in any $n$-box with $J \subset \omega$, since any such box has $n$ values in $U_n$. Thus, $s$ is not in any set in $F_0$, so it is not a cover. QED

In particular, to answer the question at the end, it suffices to take any uncountable $\kappa$.

The answer is Yes.

Theorem. The following are equivalent for any Hausdorff space X.

1. X is compact.

2. X^κ is Lindelöf for any cardinal κ.

3. X^ω₁ is Lindelöf.

Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if X is compact, then X^κ is compact and hence Lindelöf.

So suppose that we have a space X that is not compact, but X^ω₁ is Lindelöf. It follows that X is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets U₀ subset U₁ subset ... U_n ... with the union U{ U_n | n in ω} = X.

For each J subset ω of size n, let U_J be the set {s in X^ω₁ | s(α) in U_n for each α in J}. As the size of J increases, the set U_J allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call U_J an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such U_J for all finite J subset ω₁.

This F is a cover of X^ω₁. To see this, consider any point s in X^ω₁. For each α in ω₁, there is some n with s(α) in U_n. Since ω₁ is uncountable, there must be some value of n that is repeated unboundedly often, in particular, some n occurs at least n times. Let J be the coordinates where this n appears. Thus, s is in U_J. So F is a cover.

Since X^ω₁ is Lindelöf, there must be a countable subcover F₀. Let J* be the union of all the finite J that appear in the U_J in this subcover. So J* is a countable subset of ω₁. Note that J* cannot be finite, since then the sizes of the J appearing in F₀ would be bounded and it could not cover X^ω₁. We may rearrange indices and assume without loss of generality that J*=ω is the first ω many coordinates. So F₀ is really a cover of X^ω, by ignoring the other coordinates.

But this is impossible. Define a sequence s in X^ω₁ by choosing s(n) to be outside U_n+1, and otherwise arbitrary. Note that s is in U_n in fewer than n coordinates below ω, and so s is not in any n-box with J subset ω, since any such box has n values in U_n. Thus, s is not in any set in F₀, so it is not a cover. QED

In particular, to answer the question at the end, it suffices to take any uncountable kappa.

The answer is Yes.

Theorem. The following are equivalent for any Hausdorff space X.

1. X is compact.

2. X^κ is Lindelöf for any cardinal κ.

3. X^ω₁ is Lindelöf.

Proof. The forward implications are easy, using Tychonoff for 1 implies 2, since if X is compact, then X^κ is compact and hence Lindelöf.

So suppose that we have a space X that is not compact, but X^ω₁ is Lindelöf. It follows that X is Lindelöf. Thus, there is a countable cover having no finite subcover. From this, we may construct a strictly increasing sequence of open sets U₀ subset U₁ subset ... U_n ... with the union U{ U_n | n in ω} = X.

For each J subset ω₁ of size n, let U_J be the set {s in X^ω₁ | s(α) in U_n for each α in J}. As the size of J increases, the set U_J allows more freedom on the coordinates in J, but restricts more coordinates. If J has size n, let us call U_J an open n-box, since it restricts the sequences on n coordinates. Let F be the family of all such U_J for all finite J subset ω₁.

This F is a cover of X^ω₁. To see this, consider any point s in X^ω₁. For each α in ω₁, there is some n with s(α) in U_n. Since ω₁ is uncountable, there must be some value of n that is repeated unboundedly often, in particular, some n occurs at least n times. Let J be the coordinates where this n appears. Thus, s is in U_J. So F is a cover.

Since X^ω₁ is Lindelöf, there must be a countable subcover F₀. Let J* be the union of all the finite J that appear in the U_J in this subcover. So J* is a countable subset of ω₁. Note that J* cannot be finite, since then the sizes of the J appearing in F₀ would be bounded and it could not cover X^ω₁. We may rearrange indices and assume without loss of generality that J*=ω is the first ω many coordinates. So F₀ is really a cover of X^ω, by ignoring the other coordinates.

But this is impossible. Define a sequence s in X^ω₁ by choosing s(n) to be outside U_n+1, and otherwise arbitrary. Note that s is in U_n in fewer than n coordinates below ω, and so s is not in any n-box with J subset ω, since any such box has n values in U_n. Thus, s is not in any set in F₀, so it is not a cover. QED

In particular, to answer the question at the end, it suffices to take any uncountable kappa.