Let $V$ be a vector space over a field $\mathbb{F}$ and $W,X\subset V$ linear subspaces. We say $$V=W+X$$ provided each $v\in V$ can be written $$v=w+x,w\in W,x\in X.$$ We say $$V=W\oplus X$$ provided each $v\in V$ has a unique representation $v=w+x$. Show that $$V=W\oplus X\iff V=W+X\,\land\,W\cap X=\{0\}.$$
$(\Longrightarrow)$ Since if $V=W\oplus X$, every $v\in V$ can be uniquely written as $v=w+x$ for vectors $w\in W$, $x\in X$, it follows immediately that $V=W\oplus X$ implies that $V=W+X$.
Assume that $V=W\oplus X$ but $W\cap X\neq\{0\}$. Then there exists some nonzero vector $\nu\in W\cap X$ such that $\nu=w^*=x^*$ for some $w^*\in W, x^*\in X$. Let $v=w^*+x$ for $v\in V,x\in X$. Then since $\nu=w^*$, we have that $v=\nu+x=0+(\nu+x)$. Since $0\in W$ (a necessary condition for $W$ to be a linear subspace of $V$) and since $\nu+x\in X$ (since $\nu$ and $x$ are vectors in $X$ and $X$ is a linear subspace of $V$, the sum $\nu+x$ must be an element of $X$), $v$ can be represented as $v=w+x$ in two different ways. By contraposition, if $V=W\oplus X$ we must have that $W\cap X=\{0\}$.
$(\Longleftarrow)$ Since by assumption $V=W+X$, let $v\in V$ such that $v=w_1+x_1,w_1\in W,x_1\in X$. Assume that there exist vectors $w_2\in W$ and $x_2\in X$ such that $v=w_2+x_2$. Then we have that $w_1+x_1=w_2+x_2$, and rearranging we find that $w_1-w_2=x_2-x_1$. We know that $w_1-w_2\in W$ and $x_2-x_1\in X$ since $W$ and $X$ are linear subspaces. However, this implies that $w_1-w_2\in W\cap X$, which contradicts our assumption that $W\cap X=\{0\}$. Therefore, $v$ must have only one representation as $v=w+x$, and therefore we have that $V=W\oplus X$. $\Box$
Is this proof correct? I am self-studying linear algebra and would like to check my proof technique regarding linear subspaces before I move on to ensure that I am on the right track.
You definitely have the right ideas and your proof is correct. I'd just suggest two small things:
1) When you're onto proving that $W\cap X = \{0\}$ and so you assume the contrary, there was no need to use more letters for $\nu$, just use $\nu$ all the way. I know that the idea was to reinforce how it's an element of both vector spaces, but it just makes it more confusing, imo, and it really doesn't achieve much. You barely even use them, and when you do, using just $\nu$ would have been just as clear, if not clearer.
2) a small correction: $w_1-w_2\in W\cap X$ doesn't really contradict the fact that $W\cap X =\{0\}$. It's just that, both facts combined imply that $w_1-w_2=0$, which leads to the same conclusion.