continuous map of metric spaces and compactness 
Let $f:X\rightarrow Y$ be a continuous map of metric spaces. Show that if $A\subseteq X$ is compact, then $f(A)\subseteq Y$ is compact.

I am using this theorem: If $A\subseteq X$ is sequentially compact, it is compact.
Also this definition: A set $A\subseteq X$ is sequentially compact if every sequence in $A$ has a convergent subsequence in $A$. 
Attempt at a proof:
Let $\{y_n\}\subseteq f(A)$. Since $f$ is continuous, $\{y_n\}=f(x_n)$ for some $\{x_n\}\subseteq A$. If  $A\subseteq X$ is compact, every sequence $\{x_n\}\subseteq A$ has a subsequence that converges to a point in $A$, say $\{x_{n_k}\}\rightarrow a\in A$. Since $f$ is continuous, $f(x_{n_k})\rightarrow f(a)\in f(A)$. Then $f(x_{n_k})\subseteq f(A)$ is a convergent sequence in $f(A)\implies f(A)$ is compact since $\{y_n\}\subseteq f(A)$ was arbitrary.

Yet another formulation for topological spaces: If $f:X \to Y$ continuous and $f(x_\iota)$ is a net in $f(X)$, then $x_\iota$ has a converging subnet, say $x_\tau \to x$. Then $f(x_\tau) \to f(x)$, hence $f(x_\iota)$ has a converging subnet, so $f(X)$ is compact. 
