If X is regular space and Y,a dense locally compact subspace, then Y is open? prove or disprove:
If X is regular space and Y,a dense locally compact subspace, then Y is open.
 A: It is a bit long for a comment allowed on the site, so I put this as an answer.
In what follows I show that depending on the meaning of being regular, your statement can be false or true (the stronger definition of regularity makes it true, while the weaker -- false). Moreover, the stronger definition of regularity is way too strong: what you really need for the statement to be true is the requirement that "every compact subset K of $X$ is closed" (let us call it the K-property) instead of "$X$ is regular".
Note that the K-property is somewhere in between the $T_1$ and the $T_2$ properties: the Hausdorff ($T_2$) property of $X$ does imply the K-property, while there are $T_1$-spaces such that some of their compact subspaces are not closed.
1) Two definitions of a regular space.
There are two common definitions of regularity. One, weaker, just requires that any closed set and a point not in it can be separated by open neighborhoods (let us call this property "the regularity condition"). The regularity condition, by itself, does not imply that the space is Hausdorff, because nothing says that singletons are closed in this space. The other definition, stronger, adds this requirement, i.e. the space is regular if the regularity condition holds and, additionally, the space is a $T_1$-space (every one-point set is closed, e.g. Munkres's definition). This also implies that the space is Hausdorff.
Since you have not clarified your definition, I had to check it for myself whether your statement holds for either definition, and it turned out that your statement holds for the second stronger definition of a regular space only, and in that case, in fact, you do not need the whole power of regularity, all you need is the K-property, which is even weaker than the Hausdorff property.
2) The weaker definition is not enough for the statement.
First, consider the following space: $X=\{a,b,c\}$, $\mathcal{T}=\{\emptyset,X,\{a,b\},\{c\}\}$. The only two proper open sets are $\{a,b\}$ and $\{c\}$, and they are also the only two proper closed sets. There are just three possible pairs of a non-empty closed set and a point outside of it, namely, $(\{a,b\},c)$, $(\{c\},a)$ and $(\{c\},b)$ and each pair can be separated. Therefore, the regularity condition holds, and it is a regular space according to the weaker definition, but it is not even a $T_0$-space, because $a$ and $b$ are topologically indistinguishable (no open set contains just one point but not the other). Now consider set $Y=\{a,c\}$. It is dense because the only point not in $A$ is $b$, and every neighborhood of $b$ contains $a$ as well. Moreover, it is (locally) compact, and even finite. Yet, $A$ is not open.
Another example, I could think of, is the real line with double origin and the topology defined on it such that the two origins always go together (that is in every open and closed set). The space satisfies the regularity condition but it is not a $T_0$-space. Note that if we exclude one origin then the remaining subspace is topologically equivalent to the real line, and, therefore, is locally compact. It is also dense, but not open.
Hence, just the regularity property (without the requirement that the space is $T_1$) is not enough for the statement to be true.
3) The stronger definition is way too strong.
So, probably, your definition of regularity is the stronger one that also implies that the space is Hausdorff. But in this case you do not need the whole power of the regularity.
The Cofinite topology on an infinite set is an example of a $T_1$ topology that is not Hausdorff. Every subset in it is compact (hence, the space does not satisfy the K-property), and every infinite subset is dense, yet there are many infinite subsets that are not open. This shows that the $T_1$ property is not enough for the statement to be true.
On the other hand, if $X$ satisfies the K-property (which is weaker than the Hausdorff $T_2$-condition) then the statement is true. Assume $Y$ is not open, there is $y\in Y$ such that every open neighborhood of $y$ has a point in $Z=X-Y$, we show that $Y$ is not locally compact at $y$. Take any compact $K\subseteq Y$ such that $y\in K$. We show that $K$ does not contain any neighborhood of $y$ in $Y$. Indeed, if $U$ is open in $X$ and $y\in U$, then, given that $U-Y$ is non-empty and $K\subseteq Y$ is closed in $X$ (here we used the K-property), $U-K$ is non-empty and open in $X$, and, therefore, contains a point $z$ in $Y$. But then $z\in U\cap Y$ but $z\not\in K$, and $U\cap Y\not\subseteq K$.
