A topological function with only removable discontinuities I've posted similar questions here and here, but no one has answered them to my satisfaction.
Suppose that $f:\mathbb{R} \to \mathbb{R}$ is such that $\lim_{y\to x}f(y)$ exists for all $x$, that is, $f$ has only removable discontinuities. Then I claim the function $g: x\mapsto \lim_{y\to x}f(y)$ is continuous (see the first link above for a proof). 
I was wondering how we might extend this to a function $f:X \to Y$ where $X$ and $Y$ are topological spaces. Suppose that besides just existing, $\lim_{y \to x}f(y)$ is unique for all $x$. (Or if we assume $Y$ is Hausdorff, we get that for free.)  Is $g:x\mapsto \lim_{y\to x}f(y)$ continuous? Here's a partial result I've gotten:
Proposition: Suppose $f$ is as above, and $Y$ has the following property: 
$$\tag{1} \text{For every open set $V$ and every point $y\in V$, there }\\ \text{ exists a neighborhood $U$ of $y$ such that $\overline{U}\subseteq V$.}$$ 
Then $g$ is continuous.
Proof: Let $V$ be open in $Y$, and let $x\in g^{-1}(V)$. Take $U \ni g(x)$ as above. Then the definition of $g$ implies that there exists a neighborhood $\mathcal{N}$ of $x$ such that $f(\mathcal{N} \setminus \{x\})\subseteq U$. Now $\lim_{z\to \xi}f(z)$ must lie in $\overline{U}$ for all $\xi \in \mathcal{N}$, so $g(\mathcal{N})\subseteq \overline{U}\subseteq V.\quad \blacksquare$
What are the simplest kinds of topological spaces that have property $(1)$? Can we do with a weaker assumption?
 A: Note that property $(1)$ is just equivalent to regularity, so we already know many examples of spaces for which the proposition holds.
A: If you require the limit to be unique (as you said, like if we assume $Y$ to be Hausdorff), then $g$ must be continuous.
First, the function $g$ is defined everywhere, since the limit of $f$ exists and is unique everywhere. Now you have to show that the limit of $g$ exitst (and is unique, but that's implied by the Hausdorff assumption) and that it matches the value of the function. Well, assume the limit does not exist at a point $x_0$. Then, 
$$\exists V\subset Y\ \text{ s.t. }\ g(x_0)\in V, \text{ and }\, \forall U\subset X \,\, \exists y_U\in U \,\,  \text{s.t. } g(y_U)\notin V\tag{1}$$
To fix ideas, if $X,Y$ were also metric spaces, this would mean
$$
\exists \varepsilon>0 \text{ s.t. } \forall \delta>0\ \exists x_\varepsilon \,\text{ s.t. }\, |x_\varepsilon-x_0|<\delta, |g(x_\varepsilon)-g(x_0)|>\varepsilon.
$$
Given this set $V$, by definition of limit (and the definition of $g$), there is $U\subset X$ s.t. $\forall x\in U$, $f(x)\in V$. However, by (1), there is $y_U\in U$ s.t. $g(y_U)\notin V$. There are two scenarios: $g(y_U)\notin \partial V$ and $g(y_U)\in \partial V$.
1) If $g(y_U)\notin \partial V$, then $g(y_U)\in (\overline{V})^c$ and, by definition of $g$ as limit of $f$, $\exists U_y\subset U$ such that $y_U\in U_y$ and $f(x)\notin V\ \forall x\in U_y$, which contradicts the fact that $\lim_{x\to x_0} f(x)$ exists.
2) If $g(y_U)\in\partial V$, then consider $W\subset V$ such that $g(y_U)\notin \overline{W}$ (I believe this is possible, because $Y$ is Hausdorff, but you may want to double check this assumption). Then, by definition of limit at $x_0$, we have $Z\subset U$ such that $\forall x\in Z, f(x)\in W$. On the other hand, because of (1), we have that there is $y_W\in Z$ such that $g(y_W)\notin V$, and therefore $g(y_W)\notin \overline{W}$. Hence, we can use part 1) to conclude.
By the same argument I think you should be able to prove that the limit as $x\to x_0$ of $g$ has to be $g(x_0)$.
Edit: the proof was correct only if $g(y_U)\notin \bar{V}$. I added a proof for the case $g(y_U)\in\partial V$.
