Tristan pointed out in the comments that my argument showing (2) $\Rightarrow$ (1) in general is faulty. Still, I think it's true for locally compact spaces. Suppose $X$ is locally compact and (1) fails, that is, $\overline{Y}$ is not compact. Then we need to find an open cover of $X$ that has no finite subcover of $Y$. Take the family of open subsets of $X$ whose closure is compact. That does it, doesn't it?

My answer to the original question follows.

Let $(x_\alpha)$ be a net in $Y$ and suppose every open cover of $X$ has a finite subcover of $Y$. For each $\alpha$ let $F_\alpha$ be the closure, in $X$, of $\{x_\beta: \beta \geq \alpha\}$, and let $U_\alpha = X \setminus F_\alpha$. If $\{U_\alpha\}$ were an open cover of $X$ then by hypothesis there would be a finite subcover of $Y$, and then by directedness there would be a single $U_\alpha$ containing $Y$, which is absurd. So $\{U_\alpha\}$ cannot be an open cover of $X$, hence there exists $x \in X$ not in any $U_\alpha$, i.e., $x \in F_\alpha$ for all $\alpha$. So $x$ is a cluster point of the net.

Tristan pointed out in the comments that my argument showing (2) $\Rightarrow$ (1) in general is faulty. Still, I think it's true for locally compact spaces. Suppose $X$ is locally compact, $Y \subseteq X$, and every open cover of $X$ has a finite subcover of $Y$. Consider the covering of $X$ by all open subsets of whose closure is compact. Since finitely many of them cover $Y$, this implies that $\overline{Y}$ is compact.

My answer to the original question follows.

Let $(x_\alpha)$ be a net in $Y$ and suppose every open cover of $X$ has a finite subcover of $Y$. For each $\alpha$ let $F_\alpha$ be the closure, in $X$, of $\{x_\beta: \beta \geq \alpha\}$, and let $U_\alpha = X \setminus F_\alpha$. If $\{U_\alpha\}$ were an open cover of $X$ then by hypothesis there would be a finite subcover of $Y$, and then by directedness there would be a single $U_\alpha$ containing $Y$, which is absurd. So $\{U_\alpha\}$ cannot be an open cover of $X$, hence there exists $x \in X$ not in any $U_\alpha$, i.e., $x \in F_\alpha$ for all $\alpha$. So $x$ is a cluster point of the net.

Tristan, how about this. Let $(x_\alpha)$ be a net in $Y$ and suppose every open cover of $X$ has a finite subcover of $Y$. For each $\alpha$ let $F_\alpha$ be the closure, in $X$, of $\{x_\beta: \beta \geq \alpha\}$, and let $U_\alpha = X \setminus F_\alpha$. If $\{U_\alpha\}$ were an open cover of $X$ then by hypothesis there would be a finite subcover of $Y$, and then by directedness there would be a single $U_\alpha$ containing $Y$, which is absurd. So $\{U_\alpha\}$ cannot be an open cover of $X$, hence there exists $x \in X$ not in any $U_\alpha$, i.e., $x \in F_\alpha$ for all $\alpha$. So $x$ is a cluster point of the net.

Tristan pointed out in the comments that my argument showing (2) $\Rightarrow$ (1) in general is faulty. Still, I think it's true for locally compact spaces. Suppose $X$ is locally compact and (1) fails, that is, $\overline{Y}$ is not compact. Then we need to find an open cover of $X$ that has no finite subcover of $Y$. Take the family of open subsets of $X$ whose closure is compact. That does it, doesn't it?

My answer to the original question follows.

Let $(x_\alpha)$ be a net in $Y$ and suppose every open cover of $X$ has a finite subcover of $Y$. For each $\alpha$ let $F_\alpha$ be the closure, in $X$, of $\{x_\beta: \beta \geq \alpha\}$, and let $U_\alpha = X \setminus F_\alpha$. If $\{U_\alpha\}$ were an open cover of $X$ then by hypothesis there would be a finite subcover of $Y$, and then by directedness there would be a single $U_\alpha$ containing $Y$, which is absurd. So $\{U_\alpha\}$ cannot be an open cover of $X$, hence there exists $x \in X$ not in any $U_\alpha$, i.e., $x \in F_\alpha$ for all $\alpha$. So $x$ is a cluster point of the net.