Image of a precompact under the action of continuous function is a precompact

Question

Suppose $(X, d_x)$ and $(Y, d_y)$ be two metric spaces and $f\colon X \to Y$ be a continuous function .

The problem is to prove that image $f(A)$ of every precompact $A \subset X$ is also a precompact.

A subset $A$ of a topological space $X$ is said to be precompact if its closure is compact.

Thanks.

Answer 1

A metric proof: in a metric space $A$ is precompact iff every sequence in $A$ has a convergent subsequence (limit in $X$). Using this:

Take $y_n = f(x_n)$ to be any sequence in $f[A]$, where we can choose all $x_n \in A$. By precompactness of $A$, there is a subsequence $(x_{n_k})_k$ of $(x_n)_n$ that converges to some $x \in X$. But then $(y_{n_k})_k$ converges to $f(x)$ by continuity.