Consider $F:\mathbb{R}^{\mathbb{R}}\to \mathbb{R}$ given by $F(f):=f(0)$. Here $\mathbb{R}^{\mathbb{R}} $ has the product topology, then $f_n\to f $ implies $F(f_n)\to F(f) $ trivially.
Now consider the compact subspace $K:=[0,1]^{\mathbb{R}} $. It can be prooved that $F(K)=\mathbb{R} $ i.e. is not compact. From this it follows that $F $ cannot be continuous.
My question is why the function is compatible with taking limits but it is not continuous? I've thought that it has something to do with the fact that $ \mathbb{R}^{\mathbb{R}}$ is not metric, but when I think about the proof of the fact Continuous $ \leftrightarrow$ Limit Continuous I kind of feel that the only thing you need is that $f(\overline{A})\subseteq \overline{f(A)} $ which is equivalent to continuity in a general topological space. In conclusion my question is why can't you translate the proof in metric spaces to the proof in $\mathbb{R}^{\mathbb{R}} $?
You can't do that because $\mathbb{R}^{\mathbb{R}}$ is NOT a first countable topological space (https://en.wikipedia.org/wiki/First-countable_space).
Anyway, in this case $F$ is continuous as pointed out in the comments.
BUT in general
Continuity can be verified on sequences only if the space is first countable! That is, only if every point has a countable base of neighborhoods. In the case of metric spaces, this works because a metric space is always first countable. Indeed for every point $x$ $\{B(x,\frac{1}{n})\}_{n \in \mathbb{N}}$ is a countable base of neighborhoods for $x$.