Let $V$ be a vectorial space over a field F and let $w_1$ and $w_2$ be subspaces of $V$. Prove that if V is the direct sum of $w_1$ and $w_2$ iff $w_1$ is the maximum with the characteristic that $w_1\cap w_2 = \{0\}$, that means, $w_1\cap w_2 = \{0\}$ and, for any subspace $w$ of $v$ if $w_1\subset w$ implies $\{0\} \subset w \cap w_2$.
That means:
$V = w_1 \oplus w_2$ if and only if $w_1\cap w_2 = \{0\}$ and $\forall w \leq V\,(w_1 \subset w \implies \{0\} \subset w \cap w_2$)
I'm thinking of proving by contradiction. Also with properties of one basis $\beta$ of $w$. Then, from that basis say how $w_1$ should be created having on mind that $w_1$ can be the span($ \beta \setminus \{x\}$) where $x \in w \cap w_2$
Let $V$ be a vector space and let $w_{2}\subset V$ be a fixed subspace. Let $$X=\{w_{1}\subset V:w_{1}\text{ is a subspace and }w_{1}\cap w_{2}=\{0\}\}.$$ Note that $u\leq v$ if and only if $u\subset v$ gives a partial order on $X$.
We want to prove that $V=w_{1}\oplus w_{2}$ if and only if $w_{1}$ is a maximal element of $X$. Note that a maximal element is not a maximum. As noted by Kavi Rama Murthy, a maximum has the property $v\geq u$ for all $u\in X$ while a maximal element has the property $v\leq u$ if and only if $u=v$.
Let $w_{1}\in X$ such that $V=w_{1}\oplus w_{2}$. Now let $w\in X$ such that $w_{1}\subset w$. Let $x\in w\subset V$. As $V=w_{1}\oplus w_{2}$ we can find $x_{1}\in w_{1}$ and $x_{2}\in w_{2}$ such that $x_{1}+x_{2}=x$. Then $x-x_{1}=x_{2}\in w$. As $w\cap w_{2}=\{0\}$ we find that $x_{2}=0$ and $x=x_{1}\in w_{1}$ hence $w_{1}=w$ and $w_{1}$ is maximal.
Now let $w_{1}$ be maximal in $X$ and suppose that $w_{1}\oplus w_{2}\subsetneq V$. Let $x\in V\setminus (w_{1}\oplus w_{2})$ and consider $w=\text{span}(w_{1},x)$. Let $y\in w\cap w_{2}$, we can find a $\lambda\in\mathbb{R}$ and $z\in w_{1}$ such that $y=\lambda x+z$. Note that $\lambda x=y-z\in w_{1}\oplus w_{2}$, and since $x\not\in w_{1}\oplus w_{2}$ we find that $\lambda=0$ and $y=z=0$. Hence $w\in X$ and $w_{1}\subset w$, but $w_{1}\neq w$ which is a contradiction.
So $V=w_{1}\oplus w_{2}$ if and only if $w_{1}$ is a maximal element of $X$.