Q:

Proof strategy/writing for change of variables

Claim: If $f(x)=g(x)$ for all $x$, then $f(x+c)=g(x+c)$ for all x.
Proof (attempt): Set $u=x-c$, and substitute $x=u+c$.
$f(x)=g(x)$ implies $f(u+c)=g(u+c)$ for all $u$. 
Because $u$ is a dummy variable, set $u=x$ and $f(x+c)=g(x+c)$.
Is this valid proof? Is it right to set $u=x-c$ and use $u=x$? If not, how can this be proven more correctly or elegantly?

A:

Your proof is valid, but I think you're making the situation more complicated than it really is. Your claim has two parts: a hypothesis, and a conclusion. The hypothesis is "for all $x$, $f(x) = g(x)$." The conclusion is "for all $x$, $f(x + c) = g(x + c)$." Both the hypothesis and conclusion discuss, separately, a claim about "all $x$", and the fact that they both use the same letter, $x$, is a red herring: you can rewrite the conclusion as "for all $y$, $f(y+c) = g(y+c)$" without changing anything.
With the above in mind, here's how I would prove the claim. Assume that, for all $x$, $f(x) = g(x)$. We wish to show that, for all $y$, $f(y+c) = g(y+c)$, so let $y$ be an arbitrary number. We know that $f(x) = g(x)$ for all numbers $x$, so in particular we know $f(y+c) = g(y+c)$ (since $y+c$ is a number). Since $y$ was arbitrary, we know that $f(y+c) = g(y+c)$ for all $y$.

