I think the following is a counterexample: take $Y=[-1,1]\times\{a,b\}$, where $[-1,1]$ has the standard topology, and $\{a,b\}$ the discrete topology, and let $X=Y/\sim$, $y\sim y'$ if and only if there exist $x\in[0,1]$ such that $y=(x,a)$ and $y'=(x,b)$, or $y=(x,b)$ and $y'=(x,a)$. That is, identify all points except $0$. Give $X$ the quotient topology
This is the interval $[-1,1]$ with "a doubled origin", a common proving ground because the space is $T_1$ but not $T_2$, but any two points other than the doubled origins can be separated by open neighborhoods. (So, in a sense, it is "almost" Hausdorff; the Hausdorff property only fails for one choice of points, and there are lots of other points around).
Since $Y$ is compact and the quotient map is continuous and onto, $X$ is compact.
For every positive integer $n$, let $\mathcal{B}_n\subseteq Y$ be the set $[-\frac{1}{n},\frac{1}{n}]\times\{a,b\}$, and let $B_n$ be the image of $\mathcal{B_n}$ in $X$; that is, $B_n$ is the interval from $-\frac{1}{n}$ to $\frac{1}{n}$, including both origins.
$B_n$ is closed, since $X-B_n = [-1,-\frac{1}{n})\cup(\frac{1}{n},1]$ is a union of two open sets. It is also connected, because $B_n$ is a union of two connected subsets (the two copies of the interval $[-\frac{1}{n},\frac{1}{n}]$ obtained by removing one of the two $0$s) and the two subsets intersect.
What is $\cap_{n=1}^{\infty}B_n$? It's a set whose only two elements are the doubled origin points. But this subset of $X$ is not connected, because $X$ is $T_1$, so there exist open neighborhoods $U$ and $V$ such that $(0,a)\in U-V$ and $(0,b)\in V-U$. So $B\subseteq U\cup V$, $U\cap B\neq\emptyset \neq V\cap B$, and $B\cap U\cap V = \emptyset$.
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.
Best Answer
Every subset of $\mathbb R^ n$ is a union of closed sets, namely, the one-point sets consisting of each one of its points.
Yet not all subsets of $\mathbb R^n$ are closed!