Proving continuous image of compact sets are compact

Let $\{V_a\}$ be an open cover of $f(X)$. Since $f$ is continuous, we know that each of the sets $f^{-1}(V_a)$ is open. Since $X$ is compact, there are finitely many indices $a_1,...,a_n$ such that $$X\subset f^{-1}(V_{a_1})\cup ...\cup f^{-1}(V_{a_n}).$$ Since $f(f^{-1}(E))\subset E$ for every $E\subset Y$, the above implies that $$f(X)\subset V_{a_1}\cup...\cup V_{a_n}.$$ Hence, $f(X)$ is compact.

Lindsay, what you need is the intermediate value theorem, its proof is given in wikipedia.

I feel using the sequence definition is far easier to understand is much more intuitive, and the proof is nice and clean. But be careful when making the statement that being closed and bounded implies compactness, because this only holds in Euclidean space (see Heine-Borel Theorem for more about this).

Let $f:M\to N$ be a continuous function and $M$ be a compact metric space. Now let $(y_n)$ be any sequence in $f(M)$ (the image of $f$). We need to show that there exists a subsequence $y_{n_{k}}$ that converges to some $y \in f(M)$ as $k\to \infty$.

We choose a sequence $(x_n)\in M$, and since $M$ is compact by definition we have that there exists a subsequence $(x_{n_{k}})$ which converges to some $p\in M$ as $k \to \infty$. Given that a continuous function preserves the convergence of sequences i.e. if $(x_n) \to p$ in $M$, then $f((x_n)) \to f(p)$ in the image, we have that $f((x_{n_{k}})) \to f(p)$. Since $f(p) \in f(M)$, we have that our image is compact and we obtain our desired result.

Hopefully this helps, and feel free to ask any more questions :).