The answer below this question: Example of a Borel measure, which is not Borel-regular provides an example of Borel-irregular measure. Here, I am asking a harder question:
Can we find a Borel measure $\mu$ (on $\mathbb R$ or on any other space) that is NOT Borel regular, but for every compact set $K,$ $\mu(K)$ is finite?
Easy examples like the counting measure are not going to work here.
On
@Mittens already gave you a reference for a general answer in the comments, but perhaps I can provide a short proof for $\mathbb{R}$. Let $\mu$ be a locally finite measure on $\mathbb{R}$ and define $I(f) = \int_{\mathbb{R}} fd \mu$ on the space of continuous compactly supported functions on $\mathbb{R}$. By the Riesz representation theorem, there exists a Borel regular measure $\nu$ such that $I(f) = \int f d \nu$. I claim $\mu = \nu$.
Let $(a,b)$ be an open interval and $K_n = [a+n^{-1}, b-n^{-1}]$; define $f_n$ to be $1$ on $K_n$, $0$ on the complement of $(a,b)$, and linear in between (so it is continuous and looks like a trapezium). Then $f_n\uparrow \chi(a,b)$, where $\chi (a,b)$ is the indicator of $(a,b)$. Thus, by monotone convergence, $$\mu (a,b) = \lim \int f_n d\mu = \lim \int f_n d\nu = \nu (a,b)$$ Since $\mu (a,b) \leq \mu [a,b] < \infty$ we observe that $\mu$ and $\nu$ agree and are finite on all bounded open intervals. But two such measures must be equal (e.g. by the monotone class theorem), so $\mu$ is in fact regular.
As I mentioned in my comments to the OP, there are no such measures on second countable locally compact Hausdorff topological spaces. It is a theorem that if a measure on the Borel sets on a l.c.H space that has a countable basis is finite on compacts, then it is regular (inner and outer). See for example Cohn, D. L., Measure Theory, 2nd edition pp.190.
Outside the realm of second countable l.c.H spaces, here is an interesting example from the gospel of Bourbaki (Integration Col 1, Chapter IV, exercise 5)
Let $X=\{0\}\times\mathbb{R}\cup\Big\{\big(\frac1n, \frac{k}{n^2}\big): n\in\mathbb{N}, k\in\mathbb{Z}\big\}$. For each $n$ and $y\in\mathbb{R}$ define $$B_n(y)=\Big\{(u,v)\in X: 0\leq u\leq \frac1n, |v-y|\leq u\Big\}$$ Considering the collection $B_n(y)$, $n\in\mathbb{N}$ as a system of neighborhoods of $(0,y)$ and any singleton $\{(1/n,k/n^2)\}$, $n\in\mathbb{N}$ and $k\in\mathbb{Z}$ as open this defines a topology $\tau$ in $X$ that is locally compact and Hausdorff. It is clear, for instance, to check that $V_n(y)$ is a compact (and open) neighborhood of $(0,y)$ for any $y\in\mathbb{R}$.
For any $A\subset \{0\}\times \mathbb{R}$ define $\mu(A)=0$, and for any singleton of the form $\{(1/n,k/n^2)\}$, $n\in\mathbb{N}$, $k\in\mathbb{Z}$ define $\mu(\{(1/n,k/n^2)\})=\frac1{n^3}$. It is easy to see that $\mu$ extends to a Borel measure (relative to the topology $\tau$) that is finite on compact sets for the topology $\tau$. Also, $$\mu(B_n(y))\leq \sum_{m\geq n}\frac{2m+1}{m^3}<\infty$$
Any open neighborhood $B_n(y)$ of $(0,y)$ contains only one point, namely $(0,y)$, in $\{0\}\times\mathbb{R}$, that is $B_n(y)\cap B_m(y)\cap\{0\}\times \mathbb{R}=\emptyset$. Thus any open neighborhood $U$ of $\{0\}\times \mathbb{R}$ contains a continuum of cones of the form $B_{n_y}(y)$, $y\in \mathbb{R}$, and $n_y\in\mathbb{N}$ (the space $(X,\tau)$ is not paracompact). From this, it can be seen that $\mu(U)=\infty$ and so, $$0=\mu(D)\neq\inf\{\mu(U):U\in \tau, \, \{0\}\times \mathbb{R}\subset U\}=\infty$$
Notice that $(X,\tau)$ fails to be separable; hence no contradiction to the results I referenced in my comments occurs.