Arrange a list in ascending order by deleting list elements

You can use DeleteDuplicates with Greater as the second argument:

DeleteDuplicates[lst, Greater]

{1, 65, 155}

Alternatively, you can use FoldList to apply Max recursively and take Union of the resulting list:

lst = {1, 65, 40, 155, 120, 122};

Union @ FoldList[Max] @ lst

{1, 65, 155}

You can also use DeleteDuplicates in place of Union

DeleteDuplicates @ FoldList[Max] @ lst

{1, 65, 155}

Update: A few additional alternatives:

ReplaceRepeated:

lst //. {a___, b_, c_, d___} /; c < b :> {a, b, d}

{1, 65, 155}

SequenceReplace + FixedPoint:

FixedPoint[SequenceReplace[{a___, b_, c_} /; c < b :> Sequence[a, b]], lst]

{1, 65, 155}

Code Golf: Inspired by lirtosiast's answer, we can shade a few bytes using operator forms:

Union@*FoldList[Max]@lst

{1, 65, 155}

FoldList[Max]/*Union@lst

{1, 65, 155}

StringLength /@ 
 {"Union@*FoldList[Max]", 
  "FoldList[Max]/*Union",
  "DeleteDuplicates[#,#>#2&]&",
  "Pick[#,#-Max~FoldList~#,0]&"}

{20, 20, 26, 27}

On Code Golf, alephalpha came up with this in 2015:

Pick[#,#-Max~FoldList~#,0]&

Written more readably:

Pick[#, # - FoldList[Max, #], 0]&

This is somewhat idiomatic because FoldList[Max, #] is the most obvious expression for running max, but some may prefer to make the vectorized Equal explicit.

Let me preface by saying kglr's answer is better. But, a working version of your code would be:

sample = {1, 65, 40, 155, 120, 122};

(i = 1;
 newSample = sample;
 Label[1];
 If[i < Length[newSample],
  If[newSample[[i]] <= newSample[[i + 1]],
   i = i + 1; Goto[1],
   newSample = Delete[newSample, i + 1]; Goto[1]
   ]])

newSample

{1, 65, 155}

Or alternatively

i = 1;
newSample = sample;
While[i < Length[newSample],
 If[newSample[[i]] <= newSample[[i + 1]],
  i = i + 1,
  newSample = Delete[newSample, i + 1];
  ]]

newSample

{1, 65, 155}