Explain it like the student just learned about atomic radii and shielding:

Normally, as you go across a period (left to right), atomic radius decreases — each new proton pulls electrons closer. Shielding by other electrons reduces this pull a little bit.

Now, for lanthanoids (La to Lu), you are filling the 4f subshell while staying in Period 6. You are adding 14 electrons into 4f orbitals as you go from La (4$f^{0}$) to Lu (4$f^{14}$). Each new 4f electron does shield the next one from nuclear charge — but 4f orbitals are deeply diffuse (they have complex angular shapes), and the shielding by 4f electrons is very poor compared to s, p, and even d electrons.

Result: The effective nuclear charge felt by the outermost electrons (which are in 5s and 5p) increases significantly across the lanthanoid series, even though electrons are being added. This pulls in all the electron shells progressively. The ionic radius of $Ln^{3+}$ shrinks steadily from $La^{3+}$ (116 pm) to $Lu^{3+}$ (85 pm).

The Knock-on Consequence: The 5d elements in Period 6 (Hf, Ta, W…) feel the full weight of this lanthanoid contraction. The 5d orbitals of Period 6 elements have contracted to the same size as the 4d orbitals of Period 5 elements (Zr, Nb, Mo…). Therefore, Zr and Hf (Group 4) have nearly identical ionic radii (Zr: 72 pm, Hf: 71 pm) and very similar chemistry — making them the hardest pair of elements to separate in the periodic table.

One-line core idea:

Poor 4f shielding causes cumulative shrinkage of lanthanoid radii, which in turn equalises the sizes of 4d and 5d congeners in every subsequent group.