why a subspace is closed?

Another approach will be to show that a finite-dimensional normed space (over $\mathbb R$ or $\mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.

Hint:

1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.

2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.

3. Find a subsequence which has a limit in $F$. What is the limit?

To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis $\{e_1,\dots,e_n\}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,\dots,c_n\in K$ we have $$\|\sum_{i=1}^nc_ie_i\|\geq r\sum_{i=1}^n|c_i|.$$ Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)\subset F$. We know that each $x_m=\sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.