The problem is not well posed. Are $X,Y$ general metric spaces? If so, what does the euclidean metric $e$ on $Y$ mean? If not, please give the precise assumptions on $X,Y.$
We have another problem: Suppose $X=Y=\mathbb R$ with the euclidean metric. Then both $f(x)=0,g(x)=x$ belong to $C(X,Y).$ But what is $d(f,g)$ supposed to be? You have it as
$$d(f,g)= \sup_{x\in\mathbb R}|f(x)- g(x)| = \sup_{x\in\mathbb R}|0-x| = \infty.$$
There is also a problem with the lemma. You have $f_n \in C(X,Y)$ but there is no condition on $f.$ So why is $d(f_n,f)$ even defined? (Note $d_n$ should be $d$ there.)
Finally, any proof that doesn't use the completeness of $Y$ is doomed. For if $Y$ is not complete, there is a Cauchy sequence $y_n$ in $Y$ that fails to converge to a point of $Y.$ Define $f_n(x)\equiv y_n$ for all $n.$ Then $f_n$ is Cauchy in $C(X,Y)$ but fails to converge to any $f\in C(X,Y).$
Right now my answer is in the form of comments/questions, I know. I need your answers to these questions to give an answer.
Added later: Here's a way to fix things: Since you repeatedly mention the euclidean metric, let's stay in that setting and suppose $X,Y$ are both subsets of some $\mathbb R^m.$ We assume that $Y$ is complete in the usual $\mathbb R^m$ metric. Define $B =B(X,Y)$ to be the set of all bounded functions from $X$ to $Y.$ For $f,g\in B,$ define
$$d(f,g)= \sup_{x\in X} |f(x)-g(x)|.$$
Now we have something that is well defined. Verify that $d$ is a metric on $B.$ Note that $f_n\to f$ in $B$ iff $f_n\to f$ uniformly on $X.$
Now define $C_B= C_B(X,Y)$ to be the set of functions in $B$ that are continuous on $X.$ The result we want to prove is
Thm: $C_B$ is a complete metric space in the $d$ metric.
Your lemma can be stated as
Lemma: If $f_n\in C_B,$ $f\in B$ and $d(f_n,f)\to 0,$ then $f\in C_B.$
Your proof then goes through.
To prove the theorem, suppose $(f_n)$ is a Cauchy sequence in $C_B.$ Then from the definition of the $d$-metric, for each $x\in X,$ $f_n(x)$ is a Cauchy sequence of points in $Y.$ And $Y$ is complete!! Thus for each $x\in X$ the limit $\lim_{n\to \infty}f_n(x)$ exists as a point in $Y.$ We can therefore define $f:X\to Y$ to be this limit at each $x\in X.$
If we can show $f\in B$ and $f_n\to f$ uniformly on $X,$ we'll be done by the lemma. This should be familiar territory. I'll leave the proof here for now, but ask if you have any questions.
It would be a good exercise to show that $||f||_{Lip_0}$ is indeed a norm. Otherwise, here are the general steps:
Assume $f_n$ is Cauchy.
1) First step is to show pointwise convergence so we can define a limit.
$$|f_m(x) - f_n(x)| = |(f_m-f_n)(x) - (f_m-f_n)(0)| \leq ||f_m-f_n||_{Lip_0} ||x||.$$
This shows that $f_m(x)$ is Cauchy and therefore has a limit, i.e., f(x).
2) It remains to show that $f$ is in $L$. It is obvious that $f(0) = 0$. It therefore only remains to show that $f$ as defined above is Lipschitz. We have:
$$|f_n(x) - f_n(y)| \leq ||f_n||_{Lip_0} ||x-y||.$$
But $f_n$ is Cauchy, hence bounded by say $M$. Hence, $$|f_n(x) - f_n(y)| \leq M ||x-y||.$$ At the limit, we have:
$$|f(x) - f(y)| \leq M ||x-y||.$$
Hence $f$ is Lipschitz, and the result is proven
Best Answer
There are a couple of errors in your arguments as pointed out in the comments. Here is a proof: given $\epsilon >0$ there exists $N$ such that $d(f_n,f_m) <\epsilon $ for $n , m>N$. This gives $$|f_n(x)-f_m(x)| <\epsilon $$ for all $x$ for $n , m>N$ $\cdots (1)$.
Hence $(f_n(x))$ is a Cauchy sequence for each $x$. Define $f(x)$ as $\lim_{n\to \infty} f_n(x)$.
By letting $m \to \infty$ in (1) we get $|f_n(x)-f(x)| \leq \epsilon $ for all $x$ for $n >N$. Similarly, by letting $m \to \infty$ in the inequality $$|(f_n(x)-f_n(y))-(f_m(x)-f_m(y))| \leq \epsilon |x-y|$$ we get $|(f_n(x)-f_n(y))-(f(x)-f(y))| \leq \epsilon |x-y|$ for $n >N$ for all $x,y$. Fixing $n=N+1$ and using the fact that $f_{N+1}$ is Lipschitz you see that $f$ is also Lipschitz. [Use triangle inequality]. Putting these two facts together we get $d(f_n,f) \leq \epsilon$ for $n >N$. This completes the proof.