Union of infinitely many closed sets If $(K_i)_{i \in \mathbb{N}}$ is a sequence of closed sets in  $\mathbb{R}^3$, then the union of these sets $\bigcup_{i=1}^\infty K_i = K_1 \cup K_2 \cup ... $ is also closed.
My idea: ($\bigcup_{i=1}^\infty K_i)^C = \bigcap_{i=1}^\infty (K_i)^C$
I.e. , that if all $K_i$ are open, then  $\bigcap_{i=1}^\infty K_i = K_1 \cap K_2 \cap..$ is open, which is wrong.
Counterexample would be 
$$K_i := \{(x,y,z) \in \mathbb{R^3}: \Big\| \begin {pmatrix} x \\ y \\ z \end{pmatrix} \Big\| < \frac{1}{i} \}$$
Is this correct?
 A: This is not true:
$$(-5,5) = \bigcup_{n=1}^\infty[-5+\tfrac{1}{n}, 5-\tfrac{1}{n}].$$
A: 
If $(K_i)_{i \in \mathbb{N}}$ is a sequence of closed sets in  $\mathbb{R}^3$, then the union of these sets $\bigcup_{i=1}^\infty K_i = K_1 \cup K_2 \cup ... $ is also closed.

The above statement is, as you suggest, false. Let $(K_i)_{i \in \mathbb{N}}$ be a sequence of closed sets in $\mathbb{R^n}$.
Then
$\cup_{i=1}^\infty K_i$ is closed iff $(\cup_{i=1}^\infty K_i)^C$ is open, that is iff $\cap_{i=1}^\infty K_i^C$ is open.
Note that each $U_i:=K_i^C$ is open, and as you suggest, you could choose the $U_i$'s so that their intersection is not open, e.g. $U_i = \{v \in \mathbb{R^n} \mid \|v\| < \frac{1}{i}\}$, $\cap_{i=1}^\infty U_i=\{0\}$.
As others have suggested, you could alternatively give a counterexample for the original statement by choosing e.g. $K_i=\{v \in \mathbb{R}^n \mid \|v\| \leq 1-\frac{1}{i}\}$ so that $\cup_{i=1}^\infty K_i=\{v \in \mathbb{R}^n \mid \|v\| < 1\}$.
P.S. In your question you used the same symbol $K_i$ to stand for two different notions (a closed and an open set). This is a very bad practice which you should avoid as it confuses the reader.
A: To add to Tomek Kania's counter example, we can use the definition of closed sets based on closure points. A closed set is one where all the closure points are in the set. That is, each point in the closed set is a] a closure point and b] there are no other closure points. 
A union of two closed sets is a closed set simply because each point in the set can use itself as a closure point. And all the closure points are included.
For an arbitrary union of closed sets to be not a closed set, the counter example cannot rely on (a). The way out is to come up with an example that does not include all the closure points, i.e. violate (b). The counter example ends up excluding 5 and -5. These are closure points because we can always draw a sphere, however small, and find that we include a point in (-5, 5). Since we are not including these two points, we have broken the requirement (b).
I am writing this explanation for somebody like me who's just started reading this subject.
A: You have
$$
\bigcup_{n=2}^\infty \left[ \frac 1n, 1- \frac1n
\right] = (0,1)
$$so I doubt it is true. When you find a counterexample, there is no need to start writing a proof.
A: You are on the wrong trace.
Generally, the union of infinity many open sets is still open, however, the union of infinity many closed sets may not closed.
