Prove that if $\: \forall_{t \in \mathbb{R}} \: \overline{\overline{A}}_t$ is equal to cardinality $c$, then $\:\overline{\overline{\bigcup_{t \in \mathbb{R}} \: A_t}}$ is also equal to cardinality $c$.
Here is my attempt: If we know that $\: \forall_{t \in \mathbb{R}} \: \overline{\overline{A}}_t$ then $\: \forall_{t \in \mathbb{R}} \exists_{f_t} \: f_t: \mathbb{R} \rightarrow A_t $ and those fuctions are surjective.
Now we can make function $\: F: \mathbb{R} \times \mathbb{R} \rightarrow \bigcup_{t \in \mathbb{R}} \: A_t$
$F(n, m) = f_n(m)$ which is also surjective.
Since we know that $\mathbb{R} \times \mathbb{R}$ is equal to cardinality then $\:\overline{\overline{\bigcup_{t \in \mathbb{R}} \: A_t}}$ has also have to be equal to cardinality.
Is my proof done right?
You need to use the axiom of choice to choose these functions, which is a particularly delicate point. Without the axiom of choice it is consistent, for example, that the union of countably many sets, each of size continuum, has cardinality strictly larger than the continuum.
If you assume the axiom of choice, and you have shown that if $A$ surjects onto $B$, and $A$ injects onto $B$, then $|A|=|B|$, then yes. This proof is correct.
If you haven't done that, you can either do that, or require that the $A_t$'s are pairwise disjoint and $f_t$ is actually a bijection, then show that $F$ is a bijection as well. Without the assumption on the $A_t$'s you need to do another round of choices, to reduce these sets a bit. Which makes it much easier to use the axiom of choice again, and show that from an injection and a surjection from $\Bbb R$, we get equicardinality.