I believe I need a DTD to define the schema and an XSLT if I want to display it in a browser and have it look "pretty". But I'm not sure what else I would need to have a well-defined XML document that can be queried using XQuery and displayed in a web browser.
For a XML document to be queryable using XQquery you do not have to define a DTD or XSD. The purpose of DTD or XSD is to define the strict structure of a XML document and to allow validation before usage.
Modern browsers interpret XML files very nicely and show a DOM tree. If enhanced formatting of XML for browser display is necessary you have to create a XSLT transformation file and then add a directive to the original XML document pointing to the XSLT file. The browser picks that directive and uses the built-in XSLT processor to obtain the output that is then interpreted by the browser.
info.xml
<?xml version="1.0" encoding="iso-8859-1"?>
<?xml-stylesheet type="text/xsl" href="info.xslt"?>
<info>
<appName>My App</appName>
<version>1.0.129</version>
<buildTime>10-09-2008 12:44:03</buildTime>
</info>
info.xslt
<?xml version="1.0" encoding="iso-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<head>
<title>Application</title>
<style type="text/css">
body { font-family: Lucida Console; }
#outer { text-align: left; }
#name {
font-weight: bold;
font-size: 1.2em;
}
#logo {
float: left;
padding-right: 20px;
padding-bottom: 200px;
}
</style>
</head>
<body>
<xsl:apply-templates select="info" />
</body>
</html>
</xsl:template>
<xsl:template match="info">
<img id="logo" src="image.png" />
<div id="outer">
<div id="name">
<xsl:value-of select="appName"/>
</div>
<div id="version">
<xsl:value-of select="version"/>
</div>
<div id="date">
<xsl:value-of select="buildTime"/>
</div>
</div>
</xsl:template>
</xsl:stylesheet>