Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$. Prove that $V$ is the direct sum of $W_{1}$ and $W_{2}$ iff each vector in $V$ can be uniquely written as $w_{1}+w_{2}$ where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$.
MY ATTEMPT
We are going to prove the implication $(\Rightarrow)$ first.
If $v\in V = W_{1}\oplus W_{2}$, then $v = w_{1} + w_{2}$, where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$, and $W_{1}\cap W_{2} = \{0\}$.
Assume that $v = w'_{1} + w'_{2}$, where $w'_{1}\in W_{1}$ and $w'_{2}\in W_{2}$. We shall prove that $w_{1} = w'_{1}$ and $w_{2} = w'_{2}$. Indeed, \begin{align*} v = w_{1} + w_{2} = w'_{1} + w'_{2} \Longrightarrow w_{1} - w'_{1} = w'_{2} - w_{2} \end{align*}
Therefore $w_{1} - w'_{1}\in W_{1}\cap W_{2}$, that is to say, $w_{1} = w'_{1}$. Similarly, we conclude that $w_{2} = w'_{2}$.
We may now prove the case $(\Leftarrow)$.
If $v\in V$, $v = w_{1} + w_{2}$ according to the given assumption, where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$.
Since such decomposition is unique, we deduce that $V = W_{1}\oplus W_{2}$.
Indeed, let us assume that $v\in W_{1}\cap W_{2}$. Therefore $v = v + 0 = 0 + v$.
But the decomposition of $v$ is unique. Thus $v = 0$.
Could someone please double-check my reasoning?
Everything in your solution is fine. I will give you an alternative solution for the $(\Leftarrow)$ part:
Suppose that $v \in W_1 \cap W_2$. Since $0 = v+(-v) \in W_1 + W_2$ and (obviously) the zero vector also can be written as $0 = 0+0 \in W_1 + W_2$, we conclude that these decompositions are the same. Hence $v = 0$.