Every closed interval in $R^1$ is closed set (check logic)

As far as I understand from your claim, you prove that every member of a closed interval is a limit point of the interval. You need to prove that every limit point of a closed interval is a point of the interval, instead.

Pick any closed interval $[a,b]$. Pick any $x$ that is not a member of $[a,b]$, hence $x < a$ or $b < x$. Assume that $x$ is a limit point of $[a,b]$, hence it is not closed. Without loss of generality, pick $b < x$. Then in the segment $(b,x)$ there are no points of $[a,b]$, contradicting $x$ being a limit point. Then there is no point which is not a member of the closed interval and is a limit point of the interval in the same time.

With your proof and this, additionally it becomes apparent that $[a,b]$ is a perfect set.

You're supposed to prove that any limit point of $I$ is within $I$, but what you do in your proof is to a-priori assume that $x\in I$ and then prove that $x$ is a limit point of $I$. Actually what you're proving is that $I$ is dense in itself, note how your proof also works for open intervals.

Besides your proof is broken as the uncountability of a $\mathbb R$ doesn't imply that at all. First it doesn't say anything about $I$ being uncountable, and even if it were one can construct such sets that are not dense in itself. Note that a closed interval as opposed to an open does not need to have infinite elements: the closed interval $[0,0]$ only contains $0$.

What you instead should have done is to start with a limit point $x$ of $I=[a,b]$. Which in turn mean that every neighborhood $(x-\epsilon, x+\epsilon)$ of $x$ intersects $I$, that there is a point $c\in I$ such that $|x-c|<\epsilon$. Now $a\le c\le b$ which means that $x<b+\epsilon$ and $x>a-\epsilon$. This has to be true for any $\epsilon>0$ which means that $a\le x\le b$ (for assume for example that $x<a$ then we would have $a-x>0$ and the assumption $x>a-\epsilon$ would fail if $\epsilon = (a-x)/2$ since then $a-\epsilon = a-a/2+x/2 = a/2+x/2 < x$).