How to build a list of elements satisfying a certain property without building a list of all elements first?

You can use Do together with Reap-Sow:

Reap[Do[If[EvenQ[i], Sow[i]], {i, 100}]][[2, 1]] // Short
(*{2, 4, 6, 8, 10, <<40>>, 92, 94, 96, 98, 100}*)

EDIT

It is interesting to compare the Reap-Sow approach with the linked lists approach.

Reap-Sow approach:

ClearAll[ReapSow];
SetAttributes[ReapSow, HoldAll];
ReapSow[expr_, cond_, spec_] := Reap[
  Do[If[cond, Sow[expr]], spec]
][[2, 1]];

Linked lists approach

ClearAll[LinkedLists];
SetAttributes[LinkedLists, HoldAll];
LinkedLists[expr_, cond_, spec_] := CompoundExpression[
 $temp = f[],
 Do[
   If[cond, $temp = f[$temp, expr]],
   spec
  ],
  List @@ Flatten[$temp]
];

here I intentionally do not introduce local variables not to fool RepeatedTiming by, e.g. Module overhead.

Internal`Bag cheat:

ClearAll[ReapSowCompiled];
SetAttributes[ReapSowCompiled, HoldAll];
ReapSowCompiled[expr_, cond_, spec_] := Compile[
  {},
  Module[
   {bag = Internal`Bag[{0}]},
   Do[If[cond, Internal`StuffBag[bag, expr]], spec];
   Internal`BagPart[bag, All]
  ],
  CompilationTarget -> "C"
];
cf = ReapSowCompiled[i, EvenQ[i], {i, 10^4}];

Tests in a fresh kernel:

(*A fresh kernel*)
mem1 = MemoryInUse[];
a = ReapSow[i, EvenQ[i], {i, 10^4}]; // RepeatedTiming // First
MemoryInUse[] - mem1
(*0.01*)
(*58992*)

(*A fresh kernel*)
mem1 = MemoryInUse[];
b = LinkedLists[i, EvenQ[i], {i, 10^4}]; // RepeatedTiming // First
MemoryInUse[] - mem1
(*0.01*)
(*410696*)

(*A fresh kernel*)
mem1 = MemoryInUse[];
c = cf[] // Rest; // AbsoluteTiming// First
MemoryInUse[] - mem1
(*0.000326986*)
(*47184*)

So, a complex tree structure of pointers seems to involve a bigger memory overhead then an optimized Reap-Sow or Compiled cheat.

As compiled variant of Sow and Reap, you can use the undocumented Internal`Bag:

cf = Compile[{{n, _Integer}},
   Module[{bag = Internal`Bag[{0}]},
    Do[If[EvenQ[i], Internal`StuffBag[bag, i]], {i, n}];
    Internal`BagPart[bag, All]
    ],
    CompilationTarget -> "C"
   ];

n = 10000000;
a = Reap[Do[If[EvenQ[i], Sow[i]], {i, n}]][[2, 1]]; // AbsoluteTiming // First
b = Rest[cf[n]]; // AbsoluteTiming // First
a == b

5.39257

0.214701

True