The first elementary but crucial property, I would think of when talking about consequences of compactness, would be the following:
Proposition $1$. Let $X$ be a compact topological set and let $(K_i)_{i\in I}$ be a family of closed sets of $X$, then:
$$\bigcap_{i\in I}K_i\neq\varnothing\iff\left(\forall J\subset I,\#J<+\infty\Rightarrow\bigcap_{j\in J}K_j\neq\varnothing\right).$$
Proof. The direct implication is obvious.
For the reverse one let us proceed by the contrapositive and assume that one has:
$$\bigcap_{i\in I}K_i=\varnothing.$$
Then, taking the complementary, one gets:
$$\bigcup_{i\in I}X\setminus K_i=X.$$
Since $X$ is compact, there exists $J\subset I$ finite such that:
$$\bigcup_{j\in J}X\setminus K_j=X.$$
Whence the result taking again the complementary. $\Box$
Remark 1. This proposition is frequently applied when $(K_i)_{i\in I}$ is a decreasing sequence of nonempty compact sets of a topological compact set, not necessarily compact itself.
Here is an explicit example where proposition $1$ is used. It is also a good example of what one can expect from compactness assumptions.
Theorem $1$. Let $E$ be a real vector space endowed with a dot product, let $K$ be a nonempty compact convex set of $E$ and let $G$ be a compact subgroup of $\textrm{GL}(E)$. Assume that $G$ stabilises $K$, then there exists $x\in K$ such that for all $g\in G$, $g(x)=x$.
The key lemma of theorem $1$ is the following:
Lemma $1$. Let $T$ be a continuous linear map of $E$ and assume that $T(K)\subset K$, then there exists $x\in K$ such that $T(x)=x$.
Proof. Let $x_0\in K$ and let define the following sequence:
$$x_{n+1}=\frac{1}{n+1}\sum_{k=0}^nT^k(x_0)\in K,$$
using compactness of $K$, one can assume that $(x_n)_n$ converges toward $x \in K$. To conclude, notice that:
$$\|T(x_n)-x_n\|=\frac{\|T^{n+1}(x_0)-x_0\|}{n+1}\leqslant\frac{\textrm{diam}(K)}{n+1}.$$
Therefore, taking $n\to+\infty$ shows that $x$ is a fixed point of $T$. Whence the result. $\Box$
Then, using proposition $1$ applied to:
$$K_u:=\{x\in K\textrm{ s.t. }u(x)=x\},$$
one sees that theorem $1$ is true if and only if lemma $1$ holds for a finite number of continuous linear maps. The rough sketch of a proof of theorem $1$ does not capture why the compactness of $G$ is important, but it is.
This result is called Kakutani's fixed point and have deep applications. Here is a modest sample:
Proposition $2.$ Let $G$ be a compact subgroup of $\textrm{GL}_n(\mathbb{R})$, then $G$ is conjugated to a subgroup of $O(n)$.
Remark $2$. The proof uses the fact that the convex hull of a compact set is again compact, which follows from Carathéodory's theorem.
Perhaps, more substantial is the following:
Theorem $2$. Let $G$ be a compact topological group, then there exists a unique Borel probability measure $\mu$ on $G$ such that for all $g\in G$ and all Borel measurable subset $A$ of $G$, one has:
$$\mu(gA)=\mu(A).$$
In particular, for all measurable map $f\colon G\rightarrow\mathbb{R}$ and all $g\in G$, one has:
$$\int_Gf(gx)\,\mathrm{d}\mu(x)=\int_Gf(x)\,\mathrm{d}\mu(x).$$
The existence of such measures is of importance in Riemannian geometry.
Proposition $3.$ Let $G$ be a compact Lie group, then $G$ can be endowed with a bi-invariant Riemannian metric i.e. such that left and right translations are isometries.
Proof. Let $\langle\cdot,\cdot\rangle$ be a dot product on $\mathfrak{g}$, let $\mu$ be the measure of theorem $2$ and let define:
$$\forall x,y\in\mathfrak{g},\langle x,y\rangle_G:=\int_{g\in G}\langle\textrm{Ad}(g)x,\textrm{Ad}(g)y\rangle\,\mathrm{d}\mu(g),$$
then $\langle\cdot,\cdot\rangle_G$ is a dot product on $\mathfrak{g}$ which is $\textrm{Ad}\colon G\rightarrow\textrm{GL}(\mathfrak{g})$ invariant and $h\in\Gamma(TG\otimes TG)$ defined by:
$$h_g:={L_g}^*\langle\cdot,\cdot\rangle_G$$
is a bi-invariant metric on $G$. Whence the result. $\Box$
Remark 3. In fact, the set of bi-invariant metric on $G$ is in bijective correspondence with the set of $\textrm{Ad}$-invariant dot product on $\mathfrak{g}$.
From proposition $3$, it is straightforward to deduce the following:
Theorem $3$. The exponential map of a compact Lie group is surjective.
Proof. Let $G$ be a compact Lie group, by proposition $3$, let $h$ be a bi-invariant metric on $G$ and let $\nabla$ be the associated Levi-Civita connection, then using the Koszul's formula, one has:
$$\forall X,Y\in{}^G\Gamma(TG),\nabla_XY=\frac{1}{2}[X,Y].$$
From there, it is easy to see that the geodesics starting from the identity element of $G$ are the one parameter subgroups of $G$. Therefore, the Lie theoretical exponential map of $G$ coincides with the Riemannian exponential of $(G,h)$. Whence the result from Hopf-Rinow theorem. $\Box$
Remark 4. In the proof, I used that an $\textrm{Ad}$-invariant bilinear form $B$ on $\mathfrak{g}$ is $\textrm{ad}$-alternate, namely:
$$B([X,Y],Z)=B(X,[Y,Z]).$$
Example $1$. The matrix exponential map from $\textrm{SO}(n)$ to the set of skew-symmetric matrices is surjective, since for matrix Lie groups, the exponential map is the usual one.
