Is the uniform limit of continuous functions continuous for topological spaces? Question: [For notations see the context given below.]

If $E$ is a topological space and $(f_n)_{n\geq1}$ is a Cauchy sequence in $\mathcal{C}\mathcal{B}(E)$, then we know $(f_n)_{n\geq1}$ converges in $\mathcal{B}(E)$, say to $f=\displaystyle\lim_{n\to\infty}f_n$. In general, is $f$ also continuous? If not, what good hypothesis can be put on $E$ so that the question has a positive answer?

Thoughts: If $E$ is a metric space, I believe the standard "$\frac{\epsilon}{3}$" proof that the uniform limit of continuous functions is continuous gives a positive answer to the question. But in general I don't know if it is possible to translate this proof in a topological context (in some topological spaces we don't have unicity of a limit of points, for example)...
Context: If $E$ is any set, then one can show
$$
\mathcal{B}(E):=\{f:E\to\mathbb{C}|f\text{ is bounded}\}
$$
is a Banach space, where the norm is $\|f\|_{\infty}:=\displaystyle\sup_{x\in E}|f(x)|$.
Also, one can show that if $(X,\|\cdot\|)$ is a Banach space and $Y\subseteq X$ is a subspace, then
$$
Y\text{ is closed in }X\implies(Y,\|\cdot\|)\text{ is a Banach space}.
$$
Let $E$ be a topological space. Put
$$
\mathcal{C}(E):=\{f:E\to\mathbb{C}|f\text{ is continuous}\},\\
\mathcal{C}\mathcal{B}(E):=\mathcal{C}(E)\cap\mathcal{B}(E).
$$
One can show $(\mathcal{C}\mathcal{B}(E),\|\cdot\|_{\infty})$ is a normed vector space. Now, in order to show that it is a Banach space, it is sufficient to show that it is a closed subspace of $\mathcal{B}(E)$, in view of the results quoted above. This is where the question arises.
 A: The continuity of the limit of a uniformly convergent sequence in $\mathcal{CB}(E)$ is seen by essentially the same proof as in the case when $E$ is a metric space.
Fix an arbitrary $x_0 \in E$. For any $\varepsilon > 0$, there is an $n(\varepsilon)$ such that $\lVert f_n - f\rVert_\infty < \varepsilon/3$ for all $n \geqslant n(\varepsilon)$. Choose an $n\geqslant n(\varepsilon)$. Since $f_{n}$ is continuous at $x_0$, there is a neighbourhood $U$ of $x_0$ such that
$$\lvert f_n(x) - f_n(x_0)\rvert \leqslant \frac{\varepsilon}{3}$$
for all $x\in U$. Then we have
$$\lvert f(x) - f(x_0)\rvert \leqslant \lvert f(x) - f_n(x)\rvert + \lvert f_n(x) - f_n(x_0)\rvert + \lvert f_n(x_0) - f(x_0)\rvert < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$$
for all $x\in U$. So for all $\varepsilon > 0$, the preimage $f^{-1}\bigl(D_\varepsilon(f(x_0))\bigr)$ of the disk of radius $\varepsilon$ around $f(x_0)$ is a neighbourhood of $x_0$, and hence $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous.
If the codomain of the functions is an arbitrary metric space instead of $\mathbb{C}$, the same proof, with $\lvert a-b\rvert$ replaced by $d(a,b)$, shows the continuity of uniform limits.
If the codomain is an arbitrary uniform space (Hausdorff or not, that makes no difference), the same proof also works with small modifications. If we know that every uniform structure is induced by a family of pseudometrics, all we need to change from the proof for arbitrary metric spaces as the codomain is that we need to add a "for every pseudometric $d$ in the family of pseudometrics defining the uniform structure" at the beginning of the proof. If we don't know that uniform structures can always be induced by pseudometrics or don't want to use that fact, we use the fact that for every entourage $W$ on the codomain, there is a symmetric entourage $V$ with $V^3 \subset W$. Then there is an $n$ such that $(f(x),f_n(x)) \in V$ for all $x\in E$, the continuity of $f_n$ ensures the existence of a neighbourhood $U$ of $x_0$ such that $(f_n(x),f_n(x_0)) \in V$ for all $x\in U$, and thus we have $(f_n(x),f_n(x_0)) \in V^3\subset W$ for all $x\in U$, which shows the continuity of $f$ in $x_0$.
The conclusion that the uniform limit - it need not even be the limit of a sequence, it can be the limit of a net/filter as well, the proof remains essentially unchanged - of continuous maps is continuous holds in all settings where the statement makes sense, that is, for maps $X\to Y$ where $X$ is a topological space and $Y$ a uniform space. Neither space needs to be Hausdorff, first countable, or have other nice properties (beyond what is implied by $Y$ being a uniform space).
