$$a \equiv b \pmod{m} \iff m \mid a-b$$
and $m \mid a-b \implies \color{red}{mr = a-b}$ for some integer $r$
Now since we already know that $dk=a,dl=m$
Now we substitute $dk$ for $a$ and $dl$ for $m$ in the above $\color{red}{red \space equation}$ to get $$dlr = dk -b$$
now we add $b$ to both sides of the equation and subtract $dlr$ from both sides to get $$\color{blue}{b = d(k -lr)}$$ and so $d \mid b$ because $k-lr$ is an integer
Now you already know that $\gcd(a,m)=d$ and we already showed that $d \mid b$ and so $d$ is a common divisor of $b$ and $m$.
so the $$\gcd(b,m) \geq d = \gcd(a,m)$$
Now to prove equality you need to show that $\gcd(b,m) \leq d$
$\textbf{Here is how} $
Let $g =\gcd(b,m)$ then $g \mid b$ and $g \mid m$ and so there exists integers $s,t$ such that $gs = b$ and $gt = m$
Now since we are given that $$a \equiv b \pmod{m}$$ then we know $m \mid a-b$ and so there exists integer $q$ such that $$ \color{brown}{mq = a-b}$$
now we substitute $m=gt,b=gs$ in the above $\color{brown}{brown \space equation}$ to get $$gtq = a-gs$$
Now we add $gs$ to both sides of the equation and subtract to get $$gtq +gs = a$$ and so $$g(tq + s) = a$$
Hence $g \mid a$ because $tq + s$ is an integer.
Now this means that $g$ is a common divisor of $a$ and $m$ and hence $$\gcd(a,m) \geq gcd(b,m)$$
Now since we showed that $$\gcd(a,m) \geq \gcd(b,m)$$ and $$\gcd(a,m) \leq \gcd(b,m)$$ then this implies that $$\color{green}{\gcd(a,m) = \gcd(b,m)}$$
Best Answer
If $d$ is a common divisor of $b$ and $m$, and $a = b + m l$, then $d$ also divides $a$, and thus divides $a$ and $m$.
By symmetry, the converse also holds, so the set of common divisors of $a, m$, and that of $b, m$ is the same, hence the two $\gcd$ are also the same.
This is basically your proof, without resorting to fractions, and without the need to go through inequalities, it's just a matter of divisibility.