Since the question was edited, my answer also requires an update.
We show that $f$ is continuous at an arbitrary $x\in\mathbb{R}^{n}$. Let $U\subset \mathbb{R}$ be an open neighborhood of $f(x)$. Note that property $(ii)$ implies that $\{f(x)\}=f\Big(\bigcap_{n=1}^{\infty} \bar{B}(x,\frac{1}{n})\Big)\overset{(ii)}{=}\bigcap_{n=1}^{\infty}f\big(\bar{B}(x,\frac{1}{n})\big),$ where $\bar{B}(x,\frac{1}{n}):=\{y\in\mathbb{R}^{n}:\|x-y\|\leq \frac{1}{n}\}$ is the closed $\frac{1}{n}$-radius ball around $x$. By property $(i)$ each $f\big(\bar{B}(x,\frac{1}{n})\big)$ is compact (since by Heine-Borel each $\bar{B}(x,\frac{1}{n})$ is), and $\{f(x)\}$ is thus the intersection of nested non-empty compact sets in a Hausdorff space. Since $U$ is the neighborhood of this intersection, there exists $k\in\mathbb{N}$ so that every member of the intersection is inside $U$ starting from $k$, i.e. $f\big(\bar{B}(x,\frac{1}{n})\big)\subset U$ for all $n\geq k$. This property is very straight-forward to prove, it is done for example here: Nested Sequence of Non-Empty Compact Sets
Now choose e.g. $\delta=\frac{1}{k+1}>0$, whence $f(B(x,\delta))\subset U$. We have shown that $f$ is continuous at $x$, whence $f$ is in fact a continuous function (since the choice of $x$ was arbitrary).
Old answer (before OP edited the question) It seems that every sequence would have that property that you mentioned.. The property $(ii)$ makes this kind of weird and the property $(i)$ plays no role at all. Someone please correct me if I overlooked something.
Let $x\in\mathbb{R}^{m}$ be arbitrary and $(x_{n})_{n=1}^{\infty}\subset \mathbb{R}^{m}$ converging to $x$. The sets $K_{i}:=\{x_{n}:n\geq i\}\cup \{x\}$ are compact for each $i\in\mathbb{N}$ as closed and bounded subsets of $\mathbb{R}^{m}$ by Heine-Borel. Moreover, $(K_{i})_{i=1}^{\infty}$ is a decreasing sequence of compact sets with $\cap_{i=1}^{\infty} K_{i}=\{x\}$. By property $(ii)$ we have $f(\{x\})=f(\cap_{i=1}^{\infty} K_{i})=\cup_{i=1}^{\infty}f(K_{i})$, so infact $f(K_{i})=\{f(x)\}$ for all $i\in\mathbb{N}$ and in particular the sequence $(f(x_{n}))_{n=1}^{\infty}$ is a constant sequence of $f(x)$ converging to $f(x)$. Hence $\lim_{n\to\infty}f(x_{n})=f(x)$, which shows that $f$ is continuous at $x$. Moreover, since the choice of $x$ was arbitrary, we have that $f$ is a continuous function.