Is it true that
$$\lim_{x \to c}{\Bigl(\lim_{y \to x}{g(y)}\Bigr)}=L \implies \lim_{y \to c}{g(y)}=L$$
(assuming $\lim_{y \to x}{g(y)}$ exists for all $x$)
The question crossed my mind when doing a problem to do with derivatives. Specifically, the problem of whether
$$\lim_{x \to c}{f^\prime(x)} = L \implies f^\prime(c) = L$$
since expanding $f^\prime$ using the definition of a derivative yields something a lot like the expression this question asks about:
$$\lim_{x \to c}{\Bigl(\lim_{y \to x}{\frac{f(y)-f(x)}{y-x}}\Bigr)}=L \implies \lim_{y \to c}{\frac{f(y)-f(c)}{y-c}}=L$$
(assuming that inner limit always exists, i.e. assuming that $f$ is differentiable for all $x$)
Your conclusion is correct. If $\lim_{y\to x} g(y) $ exists for all $x$ then the function $G(x) =\lim_{y\to x} g(y) $ is continuous for all $x$. Now your hypothesis is $\lim_{x\to c} G(x) =L$ and by continuity of $G$ this means $G(c) =L$ which is same as saying $\lim_{y\to c} g(y) =L$.
BTW, this does not apply to the case of derivatives because then you don't have a function of type $g(y) $. The ratio $$\frac{f(y) - f(x)} {y-x} $$ can not be conceived as a function of just $y$ as the variable $x$ is also present in the ratio.
The right approach to deal with the result on derivatives is mean value theorem and you should try that.