Suppose that $V$ is a vector space, and $W$ is a subspace of $V$. If $V$ is finite dimensional, then prove $W$ too must be finite dimensional.
It seems intuitively obvious that the dimension of $W$ must be less than or equal to that of $V$, but I am uncertain of how specifically to show this to be true analytically.
This question has been asked before (seen here). However, there is not a great degree of explanation in the given answers. The answer which labels itself as a hint is most similar to my thoughts towards an approach to proving this particular theorem.
I was thinking that if $\exists n\in\mathbb{R}:n=\text{dim}(V) \implies \exists B = \{\vec{v}_1,...,\vec{v}_n\}$, a basis for $V$, with $n$ vectors $\vec{v}_1, ..., \vec{v}_1 \in V$. Since $W$ is a subspace of $V \implies \beta$, a basis for $W$, which has $m$ vectors in it can be constructed by removing $k\in\mathbb{R}$ vectors from $B$ such that $B\setminus\{\vec{v}_i,...,\vec{v}_j\}=\beta$ where $\vec{v}_i,...,\vec{v}_j$ are those $k$ vectors. This means a basis for $W$ has $m=n-k$ vectors in it, and by the definition of bases $\implies \text{dim}(W)=m$, and since $n-k \in \mathbb{R}$ and $n-k=m \implies m\in\mathbb{R}$. Thus $W$ is finite. $\square$
It seems like there may be some fault with this argument, though.
This proof doesn't work : for example taking the canonical basis $\mathscr{C}=\big((1,0),(0,1)\big)$ of $V=\mathbb{R}^2,$ you can't extract from $\mathscr{C}$ a correct basis of $W=vect(1,1).$ The good argument would be to say that if $V$ is of dimension $n$ and if $W$ was not a finite-dimensional space, then you can take from $W$ an independant family of $n+1$ vectors which would always be independant in $V,$ leading you to a contradiction.