You are right. First observation is that the completeness of the space seems important. It's easy to see that the conclusion is false in the space $\Bbb Q$. Namely the whole space is a perfect subset.
A little enhanced proof can show that this is true in any complete metric space (at the end of the answer I provide the second proof based on Baire theorem). Consider a perfect set $S\subset X$ where $X$ is a complete metric space. Since $S$ is perfect, it's closed so it's itself a complete metric space (as a closed subset of a metric space). Therefore we can proceed without supspace $X$, that is we can assume $S$ is the whole space.
First proof
Observation: any nonempty open subset of $S$ (I recall that we consider the relative topology) is infinite. Especially, all balls have got infinitely many points.
Now observe that for any ball $B(x,r)$ and any point $z\in S$ the set $B(x,r)\setminus\{z\}$ is nonempty and open, so for any $n\in \Bbb N$ there is a ball $B(y,r')$ with $r'\leq 1/n$ such that $B(y,r')\subset \overline{B(y,r')}\subset B(x,r)\setminus\{z\}$. To make the proof easy to proceed let's denote the family of such balls $B(y,r')$ as $\mathcal I_n(B(x,r),z)$.
Assume $S$ is countable: $S=\{x_1,x_2,x_3,\ldots\}$.
- Define $V_1 := B(x_1,1)$.
- Let $V_2$ be any element of $\mathcal I_2(V_1,x_1)$.
- Let $V_3$ be any element of $\mathcal I_3(V_2,x_2)$.
- Let $n\in\Bbb N$ and assume we have defined sets $V_1,V_2,\ldots,V_n$. Let $V_{n+1}$ be any element of $\mathcal I_{n+1}(V_n,x_n)$.
Mathematical induction asserts that the sequence of $V_n$ is well defined for all $n\in \Bbb N$. This sequence satisfies the condition $x_n\notin V_{n+1}$.
Now we define $F_n:=\overline{V_n}$. We see that:
- $F_1\supset F_2\supset F_3\supset\cdots$.
- $\mathrm{diam}{F_{n}}=\mathrm{diam}{V_{n}}\leq 2\cdot \frac 1{n}$, so $\mathrm{diam} F_n\to 0$.
- All the sets $F_n$ are closed and nonempty.
Therefore the set $F:=\bigcap_{n=1}^\infty F_n$ is nonempty, that is $x_m\in F\subset F_{m+1}$ for some $m\in\Bbb N$. On the other hand, from the construction we know that $x_m\notin V_{m+1}\subset F_{m+1}$. A contradiction.
This proof is very similar to the proof of Baire Theorem. This suggests that we can use this theorem to prove our fact. This led me to another proof:
Second proof
Let $S$ be a countable complete space $S=\{x_1,x_2,x_3,\ldots\}$. If $S$ is perfect then all the closed sets $\{x_n\}$ for $n\in\Bbb N$ have got empty interiors. Therefore their sum $S=\bigcup_{n=1}^\infty\{x_n\}$ has also got empty interior, which is impossible, as $\mathrm{int}\,S=S$.
Best Answer
In a metric space, a subset $E \subseteq X$ is closed if and only if whenever $x_n$ is a sequence contained in $E$ that converges to $x$ in $X$, then $x \in E$. (If you haven't seen this before, you should have a go proving it: the standard method is by contrapositive, assuming there's a sequence $x_n \in E$ that converges to $x \notin E$ and showing that $x$ is a point that makes the complement of $E$ not open.) This is where the proof uses the fact that the sets $E_n$ are closed: this statement isn't true for non-closed $E$.
For each $E_n$, the sequence $(x_m: m\geq n)$ is a convergent sequence contained in the closed set $E_n$, so the limit $x$ is in $E_n$ because it's closed. This shows that $x$ is in all $E_n$, so it's in the intersection.