Let H be a surface in Cn, and let Y be a closed irreducible set of dimension t. If Y contains H, Y equals H, or it has dimension n, and is the entire space. Let's set this case aside.
If H contains Y then H∩Y = Y. Let's set this case aside. We will prove the resulting intersection has dimension t-1.
Let R be the ring of polynomials K[x1,x2,…xn], and let P be the prime ideal in R that defines Y. Let w be the irreducible polynomial in R that produces the surface H. The intersection vanishes on the ideal generated by w and P.
The ring R/P is an integral domain with dimension t. Let b be the image of w, so that b generates the image of the principal ideal w*R. Since R/P is noetherian, there are no prime ideals strictly between 0 and b. Pulling back to R, there are no prime ideals strictly between P and P+w.
The intersection of H and Y is a closed noetherian space and can be separated into components. Let Q be a prime ideal that defines one of these components. Thus Q contains P and w, and is bounded away from P.
Suppose Q is not minimal, subject to containing P and w. A smaller prime implies a larger closed irreducible set in the intersection. Since each component is maximal, this is a contradiction. Each Q is minimal, and has dimension t-1. The intersection is therefore a union of varieties, each having dimension t-1.
You may be wondering about the intersection of a plane and a disjoint sphere. This would seem to be empty; hardly a shape of dimension 1. However, you're looking at real space, and R is not an algebraiclly closed field. In 3 dimensional complex space, a sphear and a plane always intersect in some kind of one dimensional curve.
Next let Y be an algebraic set of dimension t, which is a union of components where at least one component has dimension t. The intersection of H with this union is the union of the individual intersections. As H intersects each component, the dimension is decreased by 1. In the "largest" component, t becomes t-1, and this carries the day. For any closed set Y of dimension t, such that none of the components of Y are contained in H, H∩Y has dimension t-1.
If the generators are g1 through gt, the resulting space is the intersection of the spaces defined by g1 through gt. Remember that R is a ufd, so split each generator into its prime factors. Thus g1 might be the product of irreducibles w1*w2*w3. The space defined by g1 is the union of the 3 surfaces defined by w1, w2, and w3. The intersection of finitely many unions is the same as the union over all possible intersections. Each intersection is now based on surfaces, as defined by irreducible polynomials. Use the above to show the dimension is at least n-t. It could be higher, if one of the surfaces wholly contains the intersection of the prior surfaces. In any case, take the union over all these intersections and the resulting set has dimension at least n-t.