$V$ a $K$-vector space has only two subspaces: $\{0_V\}$ and itself, $V$. Prove that the dimension of $V$ is $\dim V=1$
We know that:
If $W\leq V$, then $\dim W\leq\dim V$, and if $\dim W=\dim V$, we know that $W=V$
My try:
First case: $W=\{0_V\}$, we know $W\leq V$ and $\dim W=0$
The basis for the previous subspace: $\beta_w=\emptyset$
We can take a vector $v$ from $V$ ($v\in V$) then $W'=\langle v\rangle$ will also be a subspace with basis $\beta_{w'}=\{v\}$, of dimension 1 ($\dim W'=1$)
We found 2 subspaces of $V$, our hypothesis is that $V$ has just 2 subspaces, then $W'=V$ and therefore $\dim V=1$.
Found this same exercise here but I don't understand the answers