Removing horizontal noise artefacts from a (SEM) image

One thing to look at is to level-adjust each line separately by dividing (or subtracting) the mean of the line's gray value:

img = ImageCrop[ColorConvert[Import["https://i.stack.imgur.com/uVv8l.jpg"], 
    "Grayscale"], {Full, 850}, {Center, Bottom}];

calib = Rescale[
   With[{mean = Mean[#]}, # - mean] & /@ ImageData[img, "Real"]
 ] // Image

This removes a good portion of the fluctuations

After that, you can try your smoothing and segmentation again

bin = DeleteSmallComponents[
  ColorNegate[MedianFilter[calib, 5]] // Binarize,
  100
];
HighlightImage[calib, bin]

I like what @halirutan did and had a slight riff on it that appears to yield a better result.

Rescale[With[{meanFiltered = MeanFilter[#, 50]}, # - meanFiltered] & /@
ImageData[img, "Real"]] // Image

The result of this:

If I get time, I might experiment a bit with the "radius" of the MeanFilter to see if that improves the results.

Update: Using Manipulate, it appears that the best results occur between about 50 and 70, after which the banding starts reappearing and is more significant at about 120.

Manipulate[calib = Rescale[With[{meanFiltered = MeanFilter[#, radius]}, # - meanFiltered] & /@ImageData[img, "Real"]] // Image, {{radius, 50}, 10, 250, 5}]

Update2: I decided to let a program find the optimal by doing this:

initialImageData = ImageData[img, "Real"];
allSortedImages = 
  With[ {choices = 
     Table[Rescale[
       With[ {meanFiltered = MeanFilter[#, radius]},
           # - 
           meanFiltered
       ] & /@ initialImageData], {radius, 30, 100, 
       1}]},
      SortBy[choices, Total[initialImageData - #] &]
  ];

That gives this as the best image.

allSortedImages[[1]] // Image

I modified this slightly just to more easily be able to find what was the optimal radius.

initialImageData = ImageData[img, "Real"];
allAdjustedImages = 
  Table[Rescale[
    With[ {meanFiltered = MeanFilter[#, radius]},
        # - meanFiltered
    ] & /@
      initialImageData], {radius, 30, 100, 1}];
allSortedImages = 
  SortBy[allAdjustedImages, Total[initialImageData - #] &];   
Position[allAdjustedImages, allSortedImages[[1]]] 

The position returned is 38 (and the image is the same as above).