Images not shown when HTML file is sent to client browser from webserver

Since you are re-inventing the wheel, i'm assuming this question is for academic purposes. If not, you can always host your css, images and other static content on a CDN and refer to them in your HTML.

Now, what you are doing here is responding to a TCP request and not a HTTP request. Your first step should be to write a Request Handler, that captures the inputstream from the TCP socket. Then parse the stream into a valid HTTP Request. Then capture the URL path from the http request and serve content based on the path. This way you can refer to relative paths on your server for images/css. The handler for those paths should be able to map to a filesystem and pick the relevant file and compose it into a http response.

I think you can do something like this in your Worker class,

    @Override
        public void run() {

            try {

           // Get html content in string or build your static string here.
           StringBuilder contentBuilder = new StringBuilder();
           try {
             BufferedReader in = new BufferedReader(new FileReader("mypage.html"));
             String str;
             while ((str = in.readLine()) != null) {
               contentBuilder.append(str);
             }
             in.close();
           } catch (IOException e) {
           }
           String content = contentBuilder.toString();

           // Here logic to search and get that file here.
           File file = new File("demo.html");

           //After getting file, to write data in that file,
           PrintWriter writer = new PrintWriter(file);
           writer.write(content);
           writer.close();

            } catch(Exception e) {
                System.out.println(e);
            }

        }

Hope this helps.

   out.println("<img src=\"https://static.oschina.net/uploads/space/2017/0730/143737_rMYR_2903254.jpg\">");

just this code can display ur image

As of Java 6, there is an http server in the JDK that you can use.

Otherwise, you need to read the request, at least the first line, and supply a different output depending on the URL in the request.

UPDATE:

You can read the HTTP line like this:

private String readHttpLine() throws IOException {
    InputStream stream = client.getInputStream();
    Reader reader = new InputStreamReader(stream);
    BufferedReader in = new BufferedReader(reader);
    String line = in.readLine();
    if (line == null) {
        throw new IOException("No HTTP line");
    }
    return line;
}

You can use a regexp to parse the HTTP line:

private static Pattern HTTP_LINE = Pattern.compile(
        "([A-Z]+) +(.*) +HTTP/([0-9]+\\.[0-9]+)",
        Pattern.CASE_INSENSITIVE);

You would then parse it like this:

        String line = readHttpLine();
        Matcher matcher = HTTP_LINE.matcher(line);
        if (!matcher.matches()) {
            throw new IOException("Invalid HTTP line");
        }
        String method = matcher.group(1).toUpperCase();
        String path = matcher.group(2);
        String httpVersion = matcher.group(3);

Now you can test the method and the path to provide the HTTP response:

        if (method.equals("GET")) {
            switch (path) {
                case "/":
                case "/index.html":
                    PrintWriter out = new PrintWriter(client.getOutputStream());       
                    out.print("HTTP/1.1 200 OK\r\n");
                    out.print("Content-type: text/html\r\n");
                    out.print("\r\n");
                    //In this line I used out.println("full html code"); but I'd like a simpler way where it can search for the html file in the directory and send it.
                    out.flush();
                    out.close();
                    break;
                case "/style.css":
                    // provide CSS stylesheet
                    break;
                case "/image1.png":
                    // provide image 1
                    break;
                case "/image2.jpeg":
                    // provide image 1
                    break;
            }
        }

Of course, unless the content is dynamic, you can keep in in files and return the appropriate file content.

UPDATE 2:

Example to supply files in resources:

...

    if (method.equals("GET")) {
        switch (path) {
            case "/":
            case "/index.html":
                Writer out = new OutputStreamWriter(
                        client.getOutputStream(), "UTF-8");       
                out.write("HTTP/1.1 200 OK\r\n");
                out.write("Content-type: text/html\r\n");
                out.write("\r\n");
                //In this line I used out.println("full html code"); but I'd like a simpler way where it can search for the html file in the directory and send it.
                out.close();
                break;
            default:
                supplyFile(path);
                break;
        }
    }

The supplyFile method looks like that:

private void supplyFile(String path) throws IOException {
    String contentType = mimetypes.getContentType(path);
    OutputStream stream = client.getOutputStream();
    Writer out = new OutputStreamWriter(stream, "UTF-8");       
    out.write("HTTP/1.1 200 OK\r\n");
    out.write("Content-type: " + contentType + "\r\n");
    out.write("\r\n");
    out.flush();
    try (InputStream istream = getClass().getResourceAsStream(path)) {
        byte[] buf = new byte[4096];
        for (int n = istream.read(buf); n > 0; n = istream.read(buf)) {
            stream.write(buf, 0, n);
        }
    }
    out.close();
}