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Suppose E is a dense subset in X, with F a proper finite subset of E. E\F is dense in X\F, but can it be the case that E\F dense in X?

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    Can $F = \emptyset$? The empty set is finite. – 2017-02-01
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    The rationals are dense in the reals, and removing some finite number of rationals should not affect density in the reals. – 2017-02-01
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    I edited your question's title because finite does not necessarily imply closed in non-Hausdorff spaces. – 2017-02-01

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It can, but it is not necessarily so, which I suspect is your actual question. Take for example any non-empty set which has the discrete topology and let $E=X$ and $F=\{x\}$ be any singleton set for $x\in E$. If you really mean just "can it ever be" then let $E=X=[0,1]$ with the usual topology inherited from $\Bbb R$ and let $F=\{0\}$ for example.

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$E\setminus F$ is dense when $X$ is a $T_1$ space and $E$ is dense and none of the points of $F$ is isolated. This holds as $X \setminus F$ is dense and open in that case. Some illustrating examples:

Some separation axiom is needed: any singleton set is dense in $X$ when it has the indiscrete (trivial) topology, and $\{0\}$ is dense in the Sierpinski space $(\{0,1\} ,\{\emptyset,\{0\}, \{0,1\}\})$, which is $T_0$ but not $T_1$.

For $X = [0,1] \cup \{2\}$ the set $E = \mathbb{Q} \cap X$ is dense in $X$ but we cannot remove $2$, as it is isolated and must be in any dense set.

The only dense set in a discrete space $X$ is $X$ itself and we cannot omit any point.