Replace special characters in XSLT

Here's a 2.0 option:

EDIT: Sorry...the 1.0 requirement was added after I started on my answer.

XML

<?xml version="1.0" encoding="UTF-8"?>
<doc>
  <Name>O'Niel</Name>
  <Name>St Peter</Name>
  <Name>A.David</Name>
</doc>

XSLT 2.0

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="*|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="text()">
    <xsl:value-of select="replace(.,'[^a-zA-Z]','')"/>
  </xsl:template>

</xsl:stylesheet>

Output

<?xml version="1.0" encoding="UTF-8"?>
<doc>
   <Name>ONiel</Name>
   <Name>StPeter</Name>
   <Name>ADavid</Name>
</doc>

Here are a couple more ways of using replace()...

Using "i" (case-insensitive mode) flag:

replace(.,'[^A-Z]','','i')

Using category escapes:

replace(.,'\P{L}','')

There is a pure XSLT way to do this.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>
    <xsl:variable name="vAllowedSymbols"
        select="'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'"/>
    <xsl:template match="node() | @*">
        <xsl:copy>
            <xsl:apply-templates select="node() | @*"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="text()">
        <xsl:value-of select="
            translate(
                .,
                translate(., $vAllowedSymbols, ''),
                ''
                )
            "/>
    </xsl:template>
</xsl:stylesheet>

Result against this sample:

<t>
    <Name>O'Niel</Name>
    <Name>St Peter</Name>
    <Name>A.David</Name>
</t>

Will be:

<t>
    <Name>ONiel</Name>
    <Name>StPeter</Name>
    <Name>ADavid</Name>
</t>