countable intersection of connected sets

Counterexample:

Consider $J_n=(\mathbb R^\times\times \mathbb R)\cup (\{0\}\times (n,\infty))$. Each $J_n$ is connected, but the intersection is $\mathbb R^\times\times\mathbb R$ which is disconnected. Basically, each $J_n$ eats away a bit more of the connection, and eventually every part of the connection gets consumed.

Edit: An edit to the question now added the condition of closed sets, which is violated by the example above. But the basic idea can still be used.

Now consider $J_n = (((-\infty,-1]\cup[1,\infty))\times\mathbb R)\cup ([-1,1]\times [n,\infty))$. It is not hard to check that those sets are closed and connected, but the intersection is $((-\infty,-1]\cup[1,\infty))\times\mathbb R$ which is disconnected.

Here is a counterexample when you remove the Hausdorff assumption. Let $X=[-1,1]\cup\{0'\}$ be topologized like the line with two origins, so neighborhoods of $0'$ are neighborhoods of $0$ but with $0$ replaced by $0'$. Note that $X$ is compact but is not Hausdorff, since $0$ and $0'$ do not have disjoint neighborhoods. Let $J_n=[-1/n,1/n]\cup\{0'\}$. Then each $J_n$ is closed and connected, but their intersection is just $\{0,0'\}$, which has the discrete topology and thus is disconnected.