Let $Y$ be a complete metric space. Then $C^0 (X,Y)$ is complete under the uniform convergence metric. I'm trying to show that the set of continuous functions $f: X \to Y$ is complete under the uniform convergence metric if $Y$ is complete.
Just to be clear, the metric is:
$$d(f,g) = \text{sup}\, \{d(f(x), g(x))\, ; \,x\in X \} $$ 
So far what I got is:
$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?
Lemma: Let $f_n \in C^0(X,Y)$, $f:X \to Y$, $\underset{n \to \infty} \lim d(f_n, f) = 0$, then $f$ is continuous. 
Proof: Take $x_0 \in X$ and $\epsilon > 0$. There exists $n_0 \in \mathbb{N}\,; \, n > n_0 \implies d(f_n, f) < \frac{\epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $\delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $\frac{\epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $\delta$, then we have:
$$d(f(x), f(x_0)) \leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) \leq \frac{3}{5}\epsilon < \epsilon$$
$\blacksquare$
We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete. 
And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f \in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.
 A: You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere. 
You have shown that if a sequence $f_n\to f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.
To do this, use completeness of $Y$ and the uniform estimate 
$$
\sup_{x\in X}|f_{n}(x)-f_m(x)|<\epsilon
$$
for $n,m\geq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.
A: 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.
A: Let’s fix it.
We assume that $X$ is a topological space, $(Y,d)$ is a complete metric space, and $C^0(X,Y)$ is a subset of a set $C(X,Y)$ of continuous functions from $X$ to $Y$. A problem (already noted in zhw’s answer) is that given $f,g\in C(X,Y)$, 
$$\bar d(f,g) = \text{sup}\, \{d(f(x), g(x))\, ; \,x\in X \} $$
(I renamed $d$ to $\bar d$, because $d$ is already used as the metric on $Y$), is not always finite, and in this case $(C(X,Y),\bar d)$ is not a metric space. There are several ways to fix it. 


*

*Put $C^0(X,Y)=C(X,Y)$ and $d’(f,g)=\min\{\bar d(f,g),1\}$ for all $f,g\in C(X,Y)$.

*If we want to keep $\bar d$ as metric and make a natural assumption that the set $C$ of all constant functions from $X$ to $Y$ belong to $C^0(X,Y)$. Then we can put 
$$C^0(X,Y)=\{f\in C(X,Y):\forall g\in C (\bar d(f,g)<\infty)\}= \{f\in C(X,Y):\exists g\in C (\bar d(f,g)<\infty)\}.$$ 


*

*We can put $C^0(X,Y)=C(X,Y)$ and allow to $\bar d$ have infinite values, and work with this generalized metric space. 


But in any of the above cases, a sequence $\{f_n\}$ of functions of $C^0(X,Y)$ is Cauchy provides that $\bar d(f_n,f_m)$ is finite for sufficiently big $n$ and $m$. 
Now, since $(Y,d)$ is a complete metric space, for each $x\in X$ there exists $f(x)=\lim_{n\to\infty} f_n(x)$. Fix any $\varepsilon>0$. There exists $N$ such that $\bar d(f_n,f_m)\le\varepsilon$ for all $n,m\ge N$. Then for each $x\in X$ we have 
$$d(f_N(x),f(x))=\lim_{m\to\infty} d(f_N(x),f_m(x))\le\varepsilon.$$ 
Thus the sequence $\{f_n\}$ converges to $f$ uniformly. Remark, that it is yet not the same that $\lim_{n\to\infty}\bar d(f_n,f)=0$, because we have not proved yet that $f\in C^0(X,Y)$, so $\bar d(f_n,f)$ may be undefined. We finish the proof as follows. 
Since the function $f_N$ is continuous, there exists a neighborhood $U$ of the point $x$ such that $d(f_N(x),f_N(x’))<\varepsilon$ for each point $x’\in U$. Then 
$$d(f(x),(x’))\le d(f(x),f_N(x))+d(f_N(x),f_N(x’))+ d(f_N(x’),f(x’))< \varepsilon+\varepsilon+\varepsilon,$$ 
so the function $f$ is continuous.
