So our professor asked us to prove that considering any subspace $S$ of a vector space $V$, the orthogonal complement $S^{\perp}$ is unique. I have devised a proof and I am not sure whether this works. Please check my proof, which is as follows:
Consider that there are at least two different orthogonal complements $S^{\perp}_1$ and $S^{\perp}_2$ of $S$. Then because we know that $V=S+S^{\perp}_1=S+S^{\perp}_2$ and the sum is direct, $v=s_1+s^{\perp}_1=s_2+s^{\perp}_2$ where $v\in V,s_1,s_2\in S,s^{\perp}_1\in S^{\perp}_1,s^{\perp}_2\in S^{\perp}_2$. $v$ is arbitrary.
So that gives $s^{\perp}_1=(s_2-s_1)+s^{\perp}_2$. Now observing that the inner product $\left<s^{\perp}_1,s_2-s_1\right>=0$ by orthogonality, and then putting in $s^{\perp}_1=(s_2-s_1)+s^{\perp}_2$ in this inner product, we obtain $||s_2-s_1||=0$ implying that $s_1=s_2=s$(say).
So we have $v=s+s^{\perp}_1=s+s^{\perp}_2$ implying $s^{\perp}_1=s^{\perp}_2$. But $s^{\perp}_1\in S^{\perp}_1$ and $s^{\perp}_2\in S^{\perp}_2$ implying that $S^{\perp}_1\subset S^{\perp}_2$ and $S^{\perp}_2\subset S^{\perp}_1$ as $v$ was arbitrary (and hence $s^{\perp}_1,s^{\perp}_2$ are arbitrary). This shows that $S^{\perp}_1=S^{\perp}_2$ and the result is proved. Further, this shows that for any vector $v=s+s^{\perp}$ the complement $s^{\perp}$ is also unique.