Find Vector Orthogonal to the Sum of Orthonormal Vectors in an Inner Product Space PREFACE: This question is more about the validity of the logic of the proof although I also propose a solution to an example problem.
Consider $\mathbb{R}^2$ endowed with Euclidean Inner Product (dot product). Let $u,v$ be orthonormal vectors in this inner product space. Find a vector orthogonal to the vector $u+v.$
Solution: Let $S$ be the set of all vectors orthogonal to the sum of two orthonormal vectors in $\mathbb{R}^2.$ Let $w=(a,b)$ be an element of $S$ with $a,b \in \mathbb{R}.$ Then $w$ must be orthogonal to $(0,1)+(1,0)=(1,1).$ Using this restriction we have that $<(1,1),(a,b)>=0,$ which implies that any vector of the form $(a,b)$ where $a=-b$ is orthogonal to the sum of any two orthonormal vectors in this inner product space. For specificity, let $w=(1,-1)$ and we are done.  $\Box$
My question is as follows: I am unsure about the proof technique used in this problem. In general terms, if I want to find an explicit object that is related to a set of objects in a specific way, is it valid to argue that if this object were related to this set of other objects, then it must be related to this specific element of this set. Then I use this fact to find the explicit form of object I am looking for. Or, was it circular to start by letting my vector be in the set of all vectors that have the desired property. In the context of the above proof, did I only find a vector orthogonal to the sum of the standard basis vectors I chose, or is my argument exhaustive with respect to the sum of all orthonormal vectors in the given IPS. Another example of this proof technique I have encountered  is in showing that the center of the ring of $n \times n$ matrices is the set of all scalar matrices. You take a matrix that commutes with all square matrices and argue that it thus must commute with certain specific matrices, and derive the desired closed form that way.
Also, if this proof technique has a specific name, please let me know as I would love to learn more/read about it.
THANK YOU
 A: This proof is invalid. If a professor wrote it, they could sprinkle a magic sentence at the start to make it true, but as you write it it is invalid.
Generally, in math, if I ask you to prove "for all $u, v$ this happens" you can't just say "take $u = (1,0), v = (0, 1)$".
BUT in linear algebra, miracles happen. And so you can perform what's called a change of basis transformation (generally in mathematics, through automorphisms one can make simplifying changes). Just as every vector in $\mathbb{R}^2$ can be written as $(a,b),$ every vector can also be written as $au + bv$ where $u, v$ are orthonormal, and the dot product of $(au+bv) \cdot (cu+dv) = ac + bd.$ In this way, you can make a unitary change of basis and assume that $u = (1, 0), v = (0, 1).$
But if you wrote this proof on an exam, I'd give you 0 marks. You need to understand when you are allowed to make simplifcations and justify them. Math is about making arguments. If you handed in an essay to an English teacher in which you were supposed to argue "The death penalty should be abolished," and handed in "Well, I don't think I deserve the death penalty" then you'd fail. This is because you've argued that a single person (you) doesn't deserve the death penalty, not that the penalty is universally wrong. And the same is true here: You solved the problem in only one case, without realizing how to solve it in general. Furthermore, you seem to think that $(1,-1)$ is literally orthogonal to $v, u,$ when really $u-v$ is orthogonal to $v, u$ and its just that in a certain basis you can write it as $(1,-1).$
The change of basis is the most fundamental idea in linear algebra in my opinion, though, and the automorphism is one of the most fundamental in all of mathematics. You need to realize how to change the structure you have to make it easier to work with--here, the structure you have is a vector space and an inner product, so you perform a unitary change of basis sending $u$ to $(1,0)$ and $v$ to $(0,1)$ to make the problem easier. Sometimes you don't need the 'unitary' part, and hence can drop the orthonormal condition. Sometimes you're working with things completely divorced from vector spaces, and need to use a different notion of automorphism (for instance, when solving a problem in geometry  that the two diagonals of a parallelogram have the same midpoint, you could notice that affine transformations preserve midpoints, and hence take an affine transformation to assume your parallelogram is the square $(0, 0), (1, 0), (0, 1), (1, 1),$ from which the claim is obvious).
A: You didn't even mention $u$ or $v$ in your proof, so how can it be correct? Your proof looks like scratch work to me, and only adresses the case when $u = e_1, v = e_2$. The technique where you assume what you want to prove, and then deduce properties is very useful in scratchwork and can be used in proofs as long as you make your assumptions clear to the reader.
To make your proof rigorous you either have to reduce the problem to the case of handling $e_1, e_2$, or just prove it directly. The most obvious way to do the reduction is to take the natural isometry $K \in L(\mathbb{R}^2)$ defined by $Ku = e_1$, $Kv = e_2$, and then use the fact that isometries preserve the inner product.
