Pick a basis $\{v_1,\dots,v_n\}$ of $V$. If $a=\sum a_iv_i$ and $b=\sum b_iv_i$, then $\Vert a-b\Vert\leq\sum\vert a_i-b_i\vert\Vert v_i\Vert$ using the triangular inequality, and if $M$ is larger than the norm of all the $v_i$s, then clearly $\Vert a-b\Vert\leq M\sum|a_i-b_i|$.
It follows that the map $T:(a_1,\dots,a_n)\in\mathbb R^n\mapsto \sum a_iv_i\in V$, which is a bijective linear map, is continuous if we endow the domain with the $1$-norm. It follows that the function $a=(a_1,\dots,a_n)\in\mathbb R^n\mapsto\Vert\sum a_iv_i\Vert$ is continuous and non-zero on the unit sphere of $\mathbb R^N$,so its minimum there is a positive number. This implies that the map $T$ has a continuous inverse.
Since $T$ and its inverse map Cauchy sequences to Cauchy sequences and limits to limits, one can use this to show that $V$ is complete.