How do linear constraints affect the convexity of my OLS-like optimisation problem? I would like to augment a linear regression (so a convex OLS problem) with some additional constraints on the coefficients to match the subject I'm working on.
Having $x\in \mathbb{R}^n$, the solution of my linear regression, and my constraints restricting $x$ with are
$$\text{low}_\text{bound} \le A.x \le \text{up}_\text{bound}.$$
$A\in \mathbb{R}^{i\times n}$, $i$ being the number of constraints I'm defining, and $\text{low}_\text{bound}, \text{up}_\text{bound} \in \mathbb{R}^i$ the bounds of my problem, respecting $\text{up}_\text{bound}-\text{low}_\text{bound}\in \mathbb{R}^{+i}$.
Are there any such constraints that would break the convexity of my problem? Or can I just say that since each constraint can be expressed as a reduction of the optimisation space to the space between two hyperplanes, I guarantee the convexity of the problem if the problem was convex at the beginning?

Your intuition is correct.
By definition, an optimization problem (that is, maximizing or minimizing the value of a function $f:\mathcal{X}\to \mathbb R$) is convex when both $f$ is a convex (or, as appropriate, a concave) function and $\mathcal X$ is a convex set in a Euclidean space $E^n.$
One definition of a convex set is that it is the intersection of a (possibly empty) collection of half-planes in $E^n.$  Each inequality in your constraints specifies a half-plane (again, by definition).  Ergo, they define a convex subset of $\mathcal X,$ which is automatically a convex set in $E^n.$
At the same time, convexity of $f$ is a global property.  Thus, the restriction of $f$ to any subset of its domain is automatically convex--there's nothing to check.
This shows that the constrained version of the original optimization problem is also convex.
