the continuous image of a compact space is compact

Consider $f:X\to Y$ a continuous and surjective function and $X$ a compact set. We will prove that $Y$ is also a compact set.

Let $\{V_a\}$ be an open covering of $Y$. By the continuity of $f$ the pre-image by $f$ of any open subset (http://planetmath.org/OpenSubset) of $Y$ will also be an open subset of $X$.So we have an open covering $\{U_a\}$ of $X$ where $U_a=f^{-1}(V_a)$.

To see this remember, since the continuity of $f$ implies that each $U_a$ is open, all we need to prove is that ${\bigcup }_a{U}_a\supset X$.Consider $x\in X$, we know that since $\{V_a\}$ is a covering of $Y$ that there exists $i$ such that $f(x)\in V_i$ but then by construction $x\in U_i$ and $\{U_a\}$ is indeed an open covering of $X$.

Since $X$ is compact we can consider a finite set of indices $\{a_i\}$ such that $\{U_{a_i}\}$ is a finite open covering of $X$, but then $\{V_{a_i}\}$ will be a finite open covering of $Y$ and it will thus be a compact set.

To see that $\{V_{a_i}\}$ is a covering of $Y$ consider $y\in Y$. By the surjectivity of $f$ there must exist (at least) one $x\in X$ such that $f(x)=y$ and since $\{U_{a_i}\}$ is a finite covering of $X$, there exists $k$ such that $x\in U_{a_k}$.But then since $f(U_{a_k})=V_{a_k}$, we must have that $y\in V_{a_k}$ and $\{V_{a_i}\}$ is indeed a finite open covering of $Y$.