Yes, you're right: Fréchet-Uryson spaces are precisely the spaces in which the sequential closure is the same as the ordinary closure. For a specific example you can use the Arens space, which is discussed in Dan Ma’s Topology Blog: the sequential closure of the set of isolated points of the Arens space is not closed.
It's true that the sequential closure operator is not a true closure operator in the usual sense of the term; this is a minor nuisance, and one might wish for a better term, but I hardly think that it qualifies as exceptionally inconvenient.
Added: To turn the sequential closure operator into a true closure operator, you have to iterate it, possibly transfinitely. That is, if $\operatorname{scl}$ denotes the sequential closure, define
$$\operatorname{scl}^\eta A=\begin{cases}
A,&\text{if }\eta=0\\\\
\operatorname{scl}\bigcup_{\xi<\eta}\operatorname{scl}^\xi A,&\text{if }\eta>0\;.
\end{cases}$$
There is always an ordinal $\eta$ such that $\operatorname{scl}^\xi A=\operatorname{scl}^\eta A$ for all $\xi\ge\eta$; if we define $\operatorname{scl}^*A=\operatorname{scl}^\eta A,$ then $\operatorname{scl}^*$ is a true closure operator, and a space $X$ is sequential iff $\operatorname{scl}_X^*$ is identical to the ordinary closure.
The reason that sequentially closed doesn't imply closed is that a sequence has too few points.
The best counterexample I can think of concerns the first uncountable ordinal, usually denoted by $\omega_1$. Under the order topology, $[0,\omega_1)\subseteq [0,\omega_1]$ is sequentially closed, but not closed.
Any convergent sequence of countable ordinals (i.e. any sequence of points in $[0,\omega_1)$) has a countable limit, so $[0,\omega_1)$ is sequentially closed. But the complement of $[0,\omega_1)$ in $[0, \omega_1]$, which is just $\{\omega_1\}$, is not open, so $[0, \omega_1)$ is not closed.
You can fix this discrepancy by using nets rather than sequences. A net is a generalisation of a sequence where you use an arbitrary set with a nice enough order as the index set, rather than limiting yourself to just $\Bbb N$.
Your flaw is thinking that because $x$ is a limit point of $F$, there is a sequence of points $x_n\in F$ converging to $F$. That's not necessarily the case (as evidenced by my example above).
Mistake in the new edit: Yes, if $F$ is sequentially closed, it contains all its sequential limit points. So of course, if $x$ is the limit of a sequence in $F$, it is contained in $F$. However, that doesn't show that $F$ is closed, because there may be limit points which aren't sequential limit points. You haven't checked those, and therefore you cannot conclude that $F$ contains them, which again means you cannot conclude that $F$ is closed.
Best Answer
Suppose that $U$ is sequentially open with respect to $\tau_{\text{seq}}$, i.e., that every sequence converging in $\tau_{\text{seq}}$ to a point of $U$ is eventually in $U$. $\langle X,\tau_{\text{seq}}\rangle$ and $\langle X,\tau\rangle$ have the same convergent sequences, so $U$ is sequentially open with respect to $\tau$ and is therefore in $\tau_{\text{seq}}$. Thus, $\langle X,\tau_{\text{seq}}\rangle$ is sequential.