There are actually a couple of weakenings to first-countability for which it could be said that "sequences suffice".
- A topological space $X$ is called Fréchet-Urysohn if for each $A \subseteq X$ and every $x \in \overline{A}$ there is a sequence in $A$ which converges to $x$.
- A topological space $X$ is called sequential if a subset $A \subseteq X$ is closed iff it contains the limits of all convergent sequences of points in $A$.
It is not too difficult to show that first-countable $\Rightarrow$ Fréchet-Urysohn $\Rightarrow$ sequential (and neither arrow reverses).
Moreover, we have the following:
Fact. Suppose $X$ is a sequential space, and $Y$ is an arbitrary topological space. Then a function $f : X \to Y$ is continuous iff for every convergent sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ in $X$ and every limit $x$ of this sequence, we have that $f(x)$ is a limit of the sequence $\langle f(x_n) \rangle_{n \in \mathbb{N}}$.
So in the larger class of Fréchet-Urysohn spaces we get the two results you mentions in the OP. (I say "every limit" and "is a limit" in the above because I am not restricting myself to Hausdorff spaces, where every convergent sequence/net has a unique limit.)
The basic idea of using net convergence in general topological spaces is to abstract away the family of (open) neighbourhoods of a point. More particularly, if given any point $x$ in a topological space $X$, he have that $$\mathcal{N}_x = \{ U \subseteq X : U\text{ is an open neighbourhood of }x \}$$ is directed by $\supseteq$. Moreover, if $A \subseteq X$, then $x \in \overline{A}$ iff $U \cap X \neq \varnothing$ for every $U \in \mathcal{N}_x$. So if you pick some $x_U \in U \cap A$ for each $U \in \mathcal{N}_x$, we get a net in $A$ converging to $x$. The next step would be to see that if you instead choose a neighbourhood base $\mathcal{B}_x$ at $x$, we still get a set which is directed by $\supseteq$, and the result still holds.
For first-countable spaces, we get the added bonus that every point has a countable descending basis $U_0 \supseteq U_1 \supseteq \cdots$, which in terms of the $\supseteq$-order is isomorphic to $\mathbb{N}$ with the usual order. This is why sequences suffice.
For Fréchet-Urysohn spaces which are not first-countable, it is not possible to work with fixed countable descending neighbourhood bases, but some other argumentation is needed. For example, in the space I describe here, it is the connection between the space and the real line (with the usual topology) that allows us to see that it is Fréchet-Urysohn.
Suppose that $f$ is discontinuous at some point $x$. Then,for some $\varepsilon>0$, if $n\in\mathbb N$, then there is a $x_n\in B\left(x,\frac1n\right)$ such that $d\bigl(f(x),f(x_n)\bigr)\geqslant\varepsilon$. Now consider the sequence$$x,x_1,x,x_2,x,x_3,\ldots$$It converges (to $x$). However, the sequence$$f(x),f(x_1),f(x),f(x_2),f(x),f(x_3),\ldots$$does not converge.
Best Answer
Two things here.
Your "counterexample" isn't a counterexample: $f(x_i) \to 14$, not 15. Indeed, if $x_i < 5$ then $2 x_i + \text{Int}(x_i) < 10 + 4 = 14$.
Secondly, even if it did converge to the right thing (15), that still wouldn't be enough to show that the function wasn't continuous. Continuous iff all sequences have $f$ tending to the right thing, and discontinuous iff some sequence has $f$ tending to the wrong thing. Showing that one sequence tends to the right thing tells you nothing.