Prove $\dim V / U$ equals $\dim V - \dim U$ without rank-nullity From Algebra by Artin:

Based on this, we see that the map from $\varphi(G) \longrightarrow G/K$ defined by $\varphi(g) \mapsto gK$ is a group isomorphism from the image (of $\varphi$) to the cosets of the kernel, so  $\varphi(G) \cong G/K$.
If $V$ is a vector space and $K$ is a subspace of $V$, call $V/K = \{ v + K : v \in V\}$ a quotient space. Under the intuitive operations $(v + K) + (u + K) = (v + u) + K$ and $\lambda(v + K) = (\lambda v) + K$, any quotient space is a vector space.
The group isomorphism above naturally extends to a vector space isomorphism $T:V \longrightarrow V'$, proving $\text{im}T \cong V/K$, where $K = \ker T$.
Now if we prove $\dim V / K = \dim V - \dim K$, the rank-nullity theorem falls out as a corollary.

Let $\pi$ be the canonical map from $V$ to $V/K$, i.e. $\pi(v) = v + K$, which is surjective with kernel $K$. The rank-nullity theorem completes the proof that $\dim V / K = \dim V - \dim K$.
But how can we prove, when $K$ is a subspace of finite dimensional $V$, that $\dim V / K = \dim V - \dim K$? WITHOUT using the rank-nullity theorem.

EDIT: to clarify, the rank-nullity theorem states that if $T:V \longrightarrow W$ and $V$ is finite dimensional, then the rank (dimension of $\text{im}T$) plus the nullity (dimension of $\ker T$) equals $\dim V$.
 A: Take a basis of $K$ it has $m=dim K$ elements. It is linearly indendent in $V$, so it can be extended to a basis in $V$ by adding $r=dim V-m$ elements $ v_1,...v_r$ Then $v_1+K,...,v_r+K$ are linearly independent in $V/K$ and span it. Hence the dim of the factor space is $r$ as claimed.
A: If $K = \left\{ \mathbf{0}_V \right\}$, where $\mathbf{0}_V$ denotes the zero vector of $V$, then $\dim K = 0$, and also
$$
V/K = \big\{ \, \{ v \} \colon v \in V \, \big\},
$$
and so
$$
\dim V/K = \dim V = \dim V - \dim K.
$$
So let us suppose that the subspace $K$ has non-zero vectors as well.
Let us suppose that $\dim K = m$, and let $\left( e_1, \ldots, e_m \right)$ be a basis (in fact an ordered basis) for $K$.
Let us suppose that $\dim V = n$.
If $K = V$, then of course
$$
V/K = \big\{ K \big\}
$$
so that
$$
\dim V/K = 0 = \dim V - \dim K.
$$
So let us suppose that $K$ is a proper subspace of $V$. Then of course $n > m$, and the ordered basis $\left( e_1, \ldots, e_m \right)$ of the subspace $K$ can be extended to an ordered  basis $\left( e_1, \ldots, e_m, e_{m+1}, \ldots, e_n \right)$ for the whole space $V$, for some vectors $e_{m+1}, \ldots, e_n \in V \setminus K$.
We now show that the (ordered) set $\left( e_{m+1} + K, \ldots, e_n + K \right)$ forms a basis (i.e. an ordered basis) for the quotient space $V/K$.
Let $v+K$ be an arbitrary element of $V/K$, where $v \in V$.
As $v \in V$ and as $\left( e_1, \ldots, e_m, e_{m+1}, \ldots, e_n \right)$ is an ordered  basis for $V$, so this $v$ can be expressed uniquely as a linear combination of the vectors $e_1, \ldots, e_m, e_{m+1}, \ldots, e_n$; that is, there exists a unique $n$-tuple $\left( \alpha_1, \ldots, \alpha_m, \alpha_{m+1}, \ldots, \alpha_n \right)$ of scalars such that
$$
v = \alpha_1 e_1 + \cdots + \alpha_m e_m + \alpha_{m+1} e_{m+1} + \cdots + \alpha_n e_n.
$$
And, as $e_1, \ldots, e_m \in K$ and as $K$ is a (vector subspace) of $V$, so we obtain
$$
\begin{align}
v+K &= \left( \alpha_1 e_1 + \cdots + \alpha_m e_m + \alpha_{m+1} e_{m+1} + \cdots + \alpha_n e_n \right) + K \\
&=  \left( \alpha_1 e_1 + K \right) + \cdots \left( \alpha_m e_m + K \right) + \left( \alpha_{m+1} e_{m+1} + K \right) + \cdots + \left( \alpha_n e_n + K \right) \\
&= \alpha_1 \left( e_1 + K \right) + \cdots \alpha_m \left( e_m + K \right) + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\
&= \alpha_1 K + \cdots + \alpha_m K + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\
&= \underbrace{K + \cdots + K}_{\mbox{$m$ terms}} + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\
&= K + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\
&= \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right).
\end{align}
$$
Note that $K$ is the so-called zero vector of the quotient (vector) space $V/K$.
Thus the ordered set $\left( e_{m+1} + K, \ldots, e_n + K \right)$ spans $V/K$.
We now show that $\left( e_{m+1} + K, \ldots, e_n + K \right)$ is linearly independent. For this suppose that, for some scalars $\beta_{m+1}, \ldots, \beta_n$, we have
$$
\beta_{m+1} \left( e_{m+1} + K \right) + \cdots \beta_n \left( e_n + K \right) = K.
$$
Note once again that $K$ is the so-called zero vector of the quotient (vector) space $V/K$. The preceding equation can be rewritten as
$$
\left( \beta_{m+1} e_{m+1} + \cdots + \beta_n e_n \right) + K = K,
$$
which implies that
$$
\beta_{m+1} e_{m+1} + \cdots + \beta_n e_n \in K,
$$
and as $\left( e_1, \ldots, e_m \right)$ is an ordered basis for $K$, so there exists a unique $m$-tuple $\beta_1, \ldots, \beta_m$ of scalars such that
$$
\beta_{m+1} e_{m+1} + \cdots + \beta_n e_n = \beta_1 e_1 + \cdots + \beta_m e_m, 
$$
which implies that
$$
 \beta_1 e_1 + \cdots + \beta_m e_m - \beta_{m+1} e_{m+1} - \cdots - \beta_n e_n  = \mathbf{0}_V,
$$
where $\mathbf{0}_V$ denotes the zero vector in $V$, and since the vectors $e_1, \ldots, e_m, e_{m+1}, \ldots, e_n$ being basis vectors are linearly independent, therefore we can conclude that
$$
\beta_1 = \cdots = \beta_m = \beta_{m+1} = \cdots = \beta_n = 0,
$$
and thus in particular we obtain
$$
\beta_{m+1} = \cdots = \beta_n = 0,
$$
thus showing the linear independence of $\left( e_{m+1} + K, \ldots, e_n + K \right)$.
Hence $\left( e_{m+1} + K, \ldots, e_n + K \right)$ is  a (an ordered) basis for $V/K$, which shows that
$$
\dim V/K = n - m = \dim V - \dim K, 
$$
as required.
A: What about using the following result:
Proposition. If $K$ is a subspace of a vector space $V$ and $V/K$ is finite-dimensional, then
$$ V \cong K \times (V/K) .$$
Proof. Let $v_{1} + K, \ldots, v_{n} + K$ be a basis for $V/K$. Then, for any $v \in V$ there exist scalars $\alpha_{1}, \ldots, \alpha_{n}$ such that
$$ v + K = \alpha_{1}(v_{1} + K) + \ldots + \alpha_{n}(v_{n} + K).$$
Now consider the linear map $\varphi: V \rightarrow K \times (V/K) $ defined by mapping $v \in V$ to
$$ \left( v - \sum_{i=1}^{n}\alpha_{i} v_{i} \hspace{0.2cm}, \hspace{0.2cm} v + K \right) .$$
This linear map is an isomorphism. $\square$
Edit 1.
Now as a corollary, suppose $V$ is finite-dimensional. Then $K$ is finite-dimensional and $V/K$ must be finite-dimensional as well, because for any basis $v_{1}, \ldots, v_{n}$ of $V$,
the list $v_{1} + K, \ldots, v_{n} + K$ generates $V/K$. Using our previous result:
$$ \dim V = \dim \left( K \times (V/K) \right) = \dim K + \dim V/K. $$
Edit 2. Let's prove that $\varphi$ is bijective. First suppose $v \in V$ is such that $\varphi(v) = ( 0_{V}, K )$ .
Notice that $0_{V}$ is the additive identity of $K$ and $K$ is the additive identity of $V/K$, so $(0_{V}, K)$ is the additive identity of $K \times (V/K)$.
By the definition of $\varphi$, it follows that
$$ v + K = K = 0 \cdot (v_{1} + K) + \ldots + 0 \cdot (v_{n} + K) ,$$
so
$$ v - \sum_{i=1}^{n} 0 \cdot v_{i} = 0_{V} $$
and $v = 0_{V}$. Hence $\ker \varphi = \{ 0_{V} \}$ and $\varphi$ is injective.
To prove surjectivity, consider an arbitrary element $(u, v + K)$ of $K \times (V/K)$. Since $V/K$ is finite-dimensional, we can write
$$ v + K = \alpha_{1}(v_{1} + K) + \ldots + \alpha_{n}(v_{n} + K).$$
Let's now take a look at the vector
$$u + \sum_{i=1}^{n} \alpha_{i} v_{i}$$
in $V$.
The equivalence class of this vector is precisely
$$ \alpha_{1}(v_{1} + K) + \ldots + \alpha_{1}(v_{n} + K) = v + K, $$
so
$$ \varphi \left( u + \sum_{i=1}^{n} \alpha_{i} v_{i} \right) = \left( u + \sum_{i=1}^{n} \alpha_{i} v_{i} - \sum_{i=1}^{n} \alpha_{i} v_{i} \hspace{0.2cm}, \hspace{0.2cm} v + K \right) = (u, v + K). $$
