Is this proof that a continuous image of a compact metric space is compact correct?

Question

I'm asked to prove that a continuous image of a compact metric space is compact, this is my attempt:

Let $A$ be compact and $f:A\to B$ be a continuous function.

Consider an arbitrary sequence $\{y_i\}\in B$.

There exists a sequence $\{x_i\}\in A \ \ \ \ s.t. \ \ \ \{ f(x_i)\} = \{y_i\}$.

Consider the subsequence $\{x_i\}_{sub}$ of $\{x_i\}$ that is convergent (must exist since $A$ is compact). Let $\{y_i\}_{sub}$ be the corresponding subsequence of $\{y_i\}$ given by $f(x_i)$ for all points in $\{x_i\}_{sub}$. Further, let $x_i,y_i$ denote the points of the subsequences rather than the sequences.

We aim to prove $"\exists N\ \ s.t \ \ D(y_i,y)<\epsilon \ \ \forall i\geq N, \ \ \epsilon > 0".$

There exists an $N\ \ s.t. \ \ D(x_i, x)<\epsilon_1 \ \ \forall i\geq N, \ \ \epsilon_1 > 0$ since $\{x_i\}_{sub}$ is convergent.

Since $f$ is continuous, $\exists \delta \ \ s.t. \ \ D(x_i,x)<\delta \implies D(f(x_i),f(x))<\epsilon.$

Pick an arbitrary $\epsilon$

There must exist $\delta$ satisfying $D(x_i,x)<\delta \implies D(f(x_i),f(x))<\epsilon$.

Take $\epsilon_1 = \delta$.

Now we have $$\exists N \ s.t. \ D(x_i, x) < \epsilon_1=\delta, \ \ \forall i\geq N$$ and thus $$\exists N \ s.t. D(f(x_i), f(x)) < \epsilon.$$ So there exists a convergent subsequence for an arbitrary sequence $\{y_i\}\in B$ and thus $B$ is compact.

Is this correct, if not, how can I modify it to be correct? If it is correct, can I improve it?

Answer 1

The last bit is the proof for a lemma, really:

If $f:X \to Y$ is continuous and $(x_n)_n \to x$ in $X$ then $f(x_n)_n \to f(x)$ in $Y$. (In words: $f$ is sequentially continuous).

Then the rest of the proof is a valid argument for the fact that the continuous image of a sequentially compact space is sequentially compact:

$(y_n)_n$ a sequence in $Y$ that we can write as $(f(x_n)_n$ for some sequence $(x_n)_n$ in $X$. The latter has a convergent subseuqence $(x_{n_k})_k$ converging to some $x \in X$. The lemma then says $(y_{n_k})_k = f(x_{n_k})_k \to f(x)$ so the sequence $(y_n)_n$ has a convergent subsequence.

Note that the above holds in all topological spaces (no metric required). The only way we use the metric in this approach is by the equivalence of sequential compactness and compactness for this class of spaces. The proofs can be made metric free and only uses open neighbourhoods e.g.

Answer 2

Consider an open cover $(U_i)_{i \in I}$ of $f(A)$. Then you can prove that $(f^{-1}(U_i))_{i \in I}$ is an open cover of $A$ (because $f$ is continuous). So if $A$ is compact, you can extract a finite cover $(f^{-1}(U_i))_{i \in I'}$ (where $I'$ is finite) from $(f^{-1}(U_i))_{i \in I}$. Just check now that $(U_i)_{i \in I'}$ is a cover of $f(A)$.