How to prove a property of sequentially compact sets The question: Let $(X, d)$ be a sequentially compact metric space. Suppose $f\colon X → R $ is a continuous
function with the property: for each $x \in X$, there exists $x′ ∈ X$ such that $\lvert f(x′)\rvert ≤ \frac{1}{2}\lvert f(x)\rvert.$  Prove that there exists a point $x_0 ∈ X$ such that $f(x_0) = 0$.
My Work: Well I started off with a sequence in X which has a subsequence which converges to a point in X, which I called $y_0$; the sequence I called it was $(y_k)_{k=1}^{\infty}$ and subsequence $(y_{k_{j}})$. Since $f$ is continuous, I have that 
$$(f(y_{k_{j}})) \rightarrow f(y_{0}).$$  
Now X is a sequentially compact metric space, which implies $f(X)$ is sequentially compact in $\mathbb{R}.$ I want to choose a subsequence which has to do with the other property that was given, but I run into a problem. I want to try to do the squeeze theorem and then use the property to find an $x_{0}$ that works. Is this the way to go? Thanks for the help! 
 A: Sequential Compactness
The definition of sequential compactness is a very simple one : A metric space $(X,d)$ is said to be sequentially compact, if for every sequence $\{x_n\} \subset X$, there is a subsequence $\{x_{n_k}\} \subset \{x_n\}$ which is convergent to some point $x \in X$ i.e. $d(x_{n_k},x) \to 0$ as $k \to \infty$.
In other words, every sequence has a convergent subsequence, and the limit (I called it $x$ above) is in $X$ itself.
Understanding the question
What is the question saying? Well, for each point in $X$, you have another point, whose function value is smaller than half the function value of the original point.
For an illustration, start with some point $y \in X$, with $f(y) = 2$ (the number $2$ is just for illustration : it could be any number. Of course, if it was zero, we would be done!)

*

*There is a $z \in X$ such that $|f(z)| \leq \frac{|f(y)|}{2} \leq 1$.


*Now, repeating with $z$, there is an $a\ in X$ such that $|f(a)| \leq \frac{|f(z)|}{2}  \leq \frac 12$.


*Now, repeating with $a$, there is a $b \in X$ such that $|f(b)| \leq \frac{|f(a)|}{2} \leq \frac 14$.


*Now, repeating with $b$ , ...
The crux of the hypothesis is therefore this : it gives us a sequence of points, whose function values are reducing to $0$!
Now, if the function values are getting closer and closer to zero, then surely there must be some point at which the value itself is equal to zero?
This does not happen all the time,  but with sequential compactness, it does.
So how do I  use sequential compactness?
Reiterating, sequential compactness means that every sequence has a convergent subsequence, and the limit is in the space itself. That is, if you give me a sequence, I will give you back a convergent subsequence, along with its limit.
Of course, the question is : which sequence do you give me, so that the subsequence and limit point I  give you back, will be useful to solve the question?
This is what I tried to indicate in my hint, and did so in the previous section : we have a sequence of points, whose function value is converging to $0$.
More formally, fix $y_1 \in X$. There exists $y_2 \in X$ such that $|f(y_2)| \leq \frac 12|f(y_1)|$. Now, repeating this with $y_2$, we get a $y_3$. More precisely, given $y_i$ for $i \geq 1$, there is a $y_{i+1}$ such that $|f(y_{i+1})| \leq \frac12 |f(y_i)|$.
Why don't you give the sequence $y_i$? Note that $|f(y_2)| \leq \frac 12 |f(y_1)|$, and $|f(y_3)| \leq \frac 12 |f(y_2)| \leq \frac 14 |f(y_1)|$. It is not difficult to see that the sequence of real numbers $f(y_i)$ converges to zero from here : I leave this to you.
Useful information after giving me $y_i$
So you gave me $y_i$ and I gave you back a subsequence $y_{i_k}$ and a point $x \in X$ such that $y_{i_k} \to x$.
Recall what our aim is : we want a point, at which the function value is zero.
Where's the point in what I gave you? The $\mathbf{x}$, obviously.

Claim : $f(x) = 0$. If this is true, we are done, since $x \in X$ and its function value is zero.

Final step
The final step requires three observations.
One is that $f$ is continuous. Therefore, if $y_{i_k} \to x$, then $f(y_{i_k}) \to f(x)$ by the sequential definition.
The second, is that every subsequence of a convergent subsequence converges to the same point. So, if $f(y_i) \to 0$, then $f(y_{i_k}) \to 0$ as well, since it is a subsequence of the convergent sequence.
The third, is that limits are unique. Note that $f(y_{i_k}) \to f(x)$ and $f(y_{i_k}) \to 0$ as well, so $f(x) = 0$.
Consequently, we complete  the problem.
