Pardox in the counter-example for a closed and bounded set that is not compact. The following question has been answered in many threads but I have a doubt about the validity of the conclusion and it almost feels like a paradox to me! I am obviously missing something as this is a standard result, and I request you to please help me understand where I am going wrong in my thought process.
Question: Consider the set $Y = \{f \in C[0,1] : ||f||_{\infty} \leq 1\}$. This set is known to be closed in $C[0, 1]$ because of the following reason.
If a sequence $f_n$ converges to $f$, with the usual definition for $||f-f_n|| \triangleq sup\{|f(x)-f_n(x)|:x \in [0,1]\}$, then it converges uniformly. As each of $f_n \in C[0,1]$, then so is $f$. The fact that each $||f_n|| \leq 1$ can be used to show that $||f|| \leq 1$ implying that $f \in Y$. It is also easy to show that $f$ is bounded.
Here comes the kicker.
Counter-example that $Y$ is not compact: Consider the sequence $f_n(x) \triangleq x^n$. This function does converge pointwise to a function $f(x) = 1$ if $x = 1$, and $0$ otherwise. But  $f \notin Y$.
My question: Isn't $f$ a limit point of the sequence $x^n$ which is a sequence of continuous functions? If so, why is this not in line with the conclusion that $Y$ is closed, as $f \notin C[0,1]$ and hence, $f \notin Y$?
My hunch: Let us look at the definition of convergence in a metric space. Let $(C[0,1], d)$ be a metric space as defined before. A sequence $(f_n) \subseteq C[0,1]$ converges to an element $f \in C[0,1]$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(f_n,f) < \epsilon$ whenever $n \geq N$.
I feel like I am getting mixed up between pointwise convergence of functions, and convergence based on the definition for a metric space of continuous functions. While it is true that the sequence of functions $f_n(x)$ converge pointwise to a function $f(x)$ on the interval $[0,1]$, they do not converge based on the metric space definition of convergence. It seems like any sequence of functions in the metric space that converges based on metric space definition of convergence, has to be uniformly convergent in the interval $[0,1]$, and our sequence $x^n$ does not converge to $f$ uniformly.
Long story short, definitions of convergence don't seem to be exactly the same which is why the sequence $x^n$, while convergent pointwise to a discontinuous function, actually does not converge at all if we base convergence on the definition of convergence in a metric space.
Please confirm if my thought process is correct and if not, please throw some light as to where I am going wrong.
 A: You wrote

It would be nice to know a bit more on the difference between the
definitions of convergence based on that of a metric space, and that
in the usual sense.

My initial response to this is that the notion of convergence goes hand in hand with the notion of metric space. Perhaps there is no notion of convergence without a metric space. (There are topological spaces in which we can talk of continuity of a function, although there need be no metric/distance whatever.)
But it occurs to me you might mean point-wise convergence when you say "in the usual sense."
In the context of your example, this means merely that we can define a function $f$ at all. So for $t_0=0.42$, we have $t_0^n \to 0$ as $n \to \infty.$ This permits us to define $f(t_0)=0,$  and so on for all $t$ in $[0,1].$ Hence $f:[0,1] \to\mathbb{R}.$
But there is a metric space operating here as well. The space is $(X,d)=(\mathbb{R,d}).$ We are saying for all $\epsilon>0$, there is a big index $N=N(\epsilon)$ beyond which $d(0,t_0^n)<\epsilon.$
